Chapter 11

71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers · 198 exercises

Problem 116

Let \(\mathrm{A}(4,-4)\) and \(\mathrm{B}(9,6)\) be points on the parabola, \(y^{2}=4 x\). Let \(\mathrm{C}\) be chosen on the arc \(\mathrm{AOB}\) of the parabola, where \(\mathrm{O}\) is the origin, such that the area of \(\triangle \mathrm{ACB}\) is maximum. Then, the area (in sq. units) of \(\triangle \mathrm{ACB}\), is: (a) \(31 \frac{1}{4}\) (b) \(30 \frac{1}{2}\) (c) 32 (d) \(31 \frac{3}{4}\)

8 step solution

Problem 117

Tangent and normal are drawn at \(\mathrm{P}(16,16)\) on the parabola \(\mathrm{y}^{2}=16 \mathrm{x}\), which intersect the axis of the parabola at A and \(\mathrm{B}\), respectively. If \(\mathrm{C}\) is the centre of the circle through the points \(\mathrm{P}, \mathrm{A}\) and \(\mathrm{B}\) and \(\angle \mathrm{CPB}=\theta\), then a value of \(\tan \theta\) is: (a) 2 (b) 3 (c) \(\frac{4}{3}\) (d) \(\frac{1}{2}\)

7 step solution

Problem 118

Tangents drawn from the point \((-8,0)\) to the parabola \(y^{2}=8 x\) touch the parabola at \(P\) and \(Q\). If \(F\) is the focus of the parabola, then the area of the triangle \(P F Q\) (in sq. units) is equal to (a) 48 (b) 32 (c) 24 (d) 64

5 step solution

Problem 119

If \(y=m x+\mathrm{c}\) is the normal at a point on the parabola \(\mathrm{y}^{2}=8 x\) whose focal distance is 8 units, then \(|\mathrm{c}|\) is equal to: (a) \(2 \sqrt{3}\) (b) \(8 \sqrt{3}\) (c) \(10 \sqrt{3}\) (d) \(16 \sqrt{3}\)

9 step solution

Problem 120

If the common tangents to the parabola, \(x^{2}=4 y\) and the circle, \(x^{2}+y^{2}=4\) intersect at the point \(P\), then the distance of \(\mathrm{P}\) from the origin, is: (a) \(\sqrt{2}+1\) (b) \(2(3+2 \sqrt{2})\) (c) \(2(\sqrt{2}+1)\) (d) \(3+2 \sqrt{2}\)

7 step solution

Problem 121

Let \(\mathrm{P}\) be the point on the parabola, \(\mathrm{y}^{2}=8 \mathrm{x}\) which is at a minimum distance from the centre \(\mathrm{C}\) of the circle, \(\mathrm{x}^{2}+(\mathrm{y}+6)^{2}=1\). Then the equation of the circle, passing through \(\mathrm{C}\) and having its centre at \(\mathrm{P}\) is: (a) \(x^{2}+y^{2}-\frac{x}{4}+2 y-24=0\) (b) \(x^{2}+y^{2}-4 x+9 y+18=0\) (c) \(x^{2}+y^{2}-4 x+8 y+12=0\) (d) \(x^{2}+y^{2}-x+4 y-12=0\)

6 step solution

Problem 122

\(\mathrm{P}\) and \(\mathrm{Q}\) are two distinct points on the parabola, \(\mathrm{y}^{2}=4 \mathrm{x}\), with parameters \(t\) and \(t_{1}\) respectively. If the normal at \(P\) passes through \(\mathrm{Q}\), then the minimum value of \(\mathrm{t}_{1}^{2}\) is : (a) 8 (b) 4 (c) 6 (d) 2

7 step solution

Problem 123

Let \(\mathrm{O}\) be the vertex and \(\mathrm{Q}\) be any point on the parabola, \(\mathrm{x}^{2}=8 \mathrm{y}\). If the point \(\mathrm{P}\) divides the line segment OQ internally in the ratio \(1: 3\), then locus of \(\mathrm{P}\) is : (a) \(y^{2}=2 x\) (b) \(x^{2}=2 y\) (c) \(x^{2}=y\) (d) \(y^{2}=x\)

5 step solution

Problem 124

Let \(\mathrm{PQ}\) be a double ordinate of the parabola, \(y^{2}=-4 x\), where P lies in the second quadrant. If R divides \(\mathrm{PQ}\) in the ratio \(2: 1\) then the locus of \(\mathrm{R}\) is : (a) \(3 y^{2}=-2 x\) (b) \(3 y^{2}=2 x\) (c) \(9 y^{2}=4 x\) (d) \(9 y^{2}=-4 x\)

6 step solution

Problem 125

The slope of the line touching both the parabolas \(y^{2}=4 x\) and \(x^{2}=-32 y\) is (a) \(\frac{1}{8}\) (b) \(\frac{2}{3}\) (c) \(\frac{1}{2}\) (d) \(\frac{3}{2}\)

8 step solution

Problem 126

A chord is drawn through the focus of the parabola \(y^{2}=6 x\) such that its distance from the vertex of this parabola is \(\frac{\sqrt{5}}{2}\), then its slope can be: (a) \(\frac{\sqrt{5}}{2}\) (b) \(\frac{\sqrt{3}}{2}\) (c) \(\frac{2}{\sqrt{5}}\) (d) \(\frac{2}{\sqrt{3}}\)

5 step solution

Problem 127

Two tangents are drawn from a point \((-2,-1)\) to the curve, \(y^{2}=4 x\). If \(\alpha\) is the angle between them, then \(|\tan \alpha|\) is equal to: (a) \(\frac{1}{3}\) (b) \(\frac{1}{\sqrt{3}}\) (c) \(\sqrt{3}\) (d) 3

5 step solution

Problem 128

Let \(\mathrm{L}_{1}\) be the length of the common chord of the curves \(\mathrm{x}^{2}+\mathrm{y}^{2}=9\) and \(\mathrm{y}^{2}=8 \mathrm{x}\), and \(\mathrm{L}_{2}\) be the length of the latus rectum of \(y^{2}=8 x\), then: (a) \(\mathrm{L}_{1}>\mathrm{L}_{2}\) (b) \(\mathrm{L}_{1}=\mathrm{L}_{2}\) (c) \(\mathrm{L}_{1}<\mathrm{L}_{2}\) (d) \(\frac{L_{1}}{L_{2}}=\sqrt{2}\)

6 step solution

Problem 129

Given : A circle, \(2 x^{2}+2 y^{2}=5\) and a parabola, \(y^{2}=4 \sqrt{5} x\). Statement-1 : An equation of a common tangent to these curves is \(y=x+\sqrt{5}\). Statement- \(2:\) If the line, \(y=m x+\frac{\sqrt{5}}{m}(m \neq 0)\) is their common tangent, then \(m\) satisfies $m^{4}-3 m^{2}+2=0 . (a) Statement- 1 is true; Statement- 2 is true; Statement-2 is a correct explanation for Statement-1. (b) Statement- 1 is true; Statement- 2 is true; Statement-2 is not a correct explanation for Statement- 1 . (c) Statement- 1 is true; Statement- 2 is false. (d) Statement- 1 is false; Statement- 2 is true.

8 step solution

Problem 130

The point of intersection of the normals to the parabola \(y^{2}=4 x\) at the ends of its latus rectum is: (a) \((0,2)\) (b) \((3,0)\) (c) \((0,3)\) (d) \((2,0)\)

5 step solution

Problem 131

Statement-1: The line \(x-2 y=2\) meets the parabola, \(y^{2}+2 x=0\) only at the point \((-2,-2)\). Statement-2: The line \(y=m x-\frac{1}{2 m}(m \neq 0)\) is tangent to the parabola, \(y^{2}=-2 x\) at the point \(\left(-\frac{1}{2 m^{2}},-\frac{1}{m}\right)\). (a) Statement- 1 is true; Statement- 2 is false. (b) Statement- 1 is true; Statement- 2 is true; Statement-2 is a correct explanation for statement- 1 . (c) Statement- 1 is false; Statement- 2 is true. (d) Statement-1 a true; Statement-2 is true; Statement-2 is not a correct explanation for statement-1.

3 step solution

Problem 132

The normal at \(\left(2, \frac{3}{2}\right)\) to the ellipse, \(\frac{x^{2}}{16}+\frac{y^{2}}{3}=1\) touches a parabola, whose equation is (a) \(y^{2}=-104 x\) (b) \(y^{2}=14 x\) (c) \(y^{2}=26 x\) (d) \(y^{2}=-14 x\)

5 step solution

Problem 133

The chord \(P Q\) of the parabola \(y^{2}=x\), where one end \(P\) of the chord is at point \((4,-2)\), is perpendicular to the axis of the parabola. Then the slope of the normal at \(Q\) is (a) \(-4\) (b) \(-\frac{1}{4}\) (c) 4 (d) \(\frac{1}{4}\)

7 step solution

Problem 134

Statement 1: \(y=m x-\frac{1}{m}\) is always a tangent to the parabola, \(y^{2}=-4 x\) for all non-zero values of \(m\). Statement 2: Every tangent to the parabola, \(y^{2}=-4 x\) will meet its axis at a point whose abscissa is non- negative. (a) Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation of Statement 1 . (b) Statement 1 is false, Statement 2 is true. (c) Statement 1 is true, Statement 2 is false. (d) Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1 .

6 step solution

Problem 135

The shortest distance between line \(y-x=1\) and curve \(x=y^{2}\) is (a) \(\frac{3 \sqrt{2}}{8}\) (b) \(\frac{8}{3 \sqrt{2}}\) (c) \(\frac{4}{\sqrt{3}}\) (d) \(\frac{\sqrt{3}}{4}\)

9 step solution

Problem 136

If two tangents drawn from a point \(P\) to the parabola \(y^{2}=4 x\) are at right angles, then the locus of \(\mathrm{P}\) is (a) \(2 x+1=0\) (b) \(x=-1\) (c) \(2 x-1=0\) (d) \(x=1\)

7 step solution

Problem 137

A parabola has the origin as its focus and the line \(x=2\) as the directrix. Then the vertex of the parabola is at (a) \((0,2)\) (b) \((1,0)\) (c) \((0,1)\) (d) \((2,0)\)

3 step solution

Problem 138

The equation of a tangent to the parabola \(y^{2}=8 x\) is \(y=x+2\). The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is (a) \((2,4)\) (b) \((-2,0)\) (c) \((-1,1)\) (d) \((0,2)\)

6 step solution

Problem 140

Let \(P\) be the point \((1,0)\) and \(Q\) a point on the locus \(y^{2}=8 x .\) The locus of mid point of \(P Q\) is (a) \(y^{2}-4 x+2=0\) (b) \(y^{2}+4 x+2=0\) (c) \(x^{2}+4 y+2=0\) (d) \(x^{2}-4 y+2=0\)

5 step solution

Problem 141

A circle touches the \(x\) - axis and also touches the circle with centre at \((0,3)\) and radius 2 . The locus of the centre of the circle is (a) an ellipse (b) a circle (c) a hyperbola (d) a parabola

6 step solution

Problem 143

The normal at the point \(\left(b t_{1}^{2}, 2 b t_{1}\right)\) on a parabola meets the parabola again in the point \(\left(b t_{2}^{2}, 2 b t_{2}\right)\), then (a) \(t_{2}=t_{1}+\frac{2}{t_{1}}\) (b) \(t_{2}=-t_{1}-\frac{2}{t_{1}}\) (c) \(t_{2}=-t_{1}+\frac{2}{t_{1}}\) (d) \(t_{2}=t_{1}-\frac{2}{t_{1}}\)

7 step solution

Problem 144

Two common tangents to the circle \(x^{2}+y^{2}=2 a^{2}\) and parabola \(y^{2}=8 a x\) are (a) \(x=\pm(y+2 a)\) (b) \(y=\pm(x+2 a)\) (c) \(x=\pm(y+a)\) (d) \(y=\pm(x+a)\)

5 step solution

Problem 145

Which of the following points lies on the locus of the foot of perpendicular drawn upon any tangent to the ellipse, \(\frac{x^{2}}{4}+\frac{y^{2}}{2}=1\) from any of its foci? (a) \((-2, \sqrt{3})\) (b) \((-1, \sqrt{2})\) (c) \((-1, \sqrt{3})\) (d) \((1,2)\)

4 step solution

Problem 146

If the normal at an end of a latus rectum of an ellipse passes through an extermity of the minor axis, then the eccentricity e of the ellipse satisfies: (a) \(\mathrm{e}^{4}+2 \mathrm{e}^{2}-1=0\) (b) \(\mathrm{e}^{2}+\mathrm{e}-1=0\) (c) \(\mathrm{e}^{4}+\mathrm{e}^{2}-1=0\) (d) \(\mathrm{e}^{2}+2 \mathrm{e}-1=0\)

5 step solution

Problem 147

If the co-ordinates of two points \(\mathrm{A}\) and \(\mathrm{B}\) are \((\sqrt{7}, 0)\) and \((-\sqrt{7}, 0)\) respectively and \(\mathrm{P}\) is any point on the conic, \(9 x^{2}+16 y^{2}=144\), then \(\mathrm{PA}+\mathrm{PB}\) is equal to: (a) 16 (b) 8 (c) 6 (d) 9

5 step solution

Problem 148

If the point \(\mathrm{P}\) on the curve, \(4 x^{2}+5 y^{2}=20\) is farthest from the point \(\mathrm{Q}(0,-4)\), then \(\mathrm{PQ}^{2}\) is equals to : (a) 36 (b) 48 (c) 21 (d) 29

6 step solution

Problem 149

Let \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b)\) bea given ellipse, length of whose latus rectum is 10 . If its eccentricity is the maximum value of the function, \(\phi(t)=\frac{5}{12}+t-t^{2}\), then \(a^{2}+b^{2}\) is equal to: (a) 145 (b) 116 (c) 126 (d) 135

11 step solution

Problem 150

Let \(x=4\) be a directrix to an ellipse whose centre is at the origin and its eccentricity is \(\frac{1}{2}\). If \(P(1, \beta), \beta>0\) is a point on this ellipse, then the equation of the normal to it at \(P\) is : (a) \(4 x-3 y=2\) (b) \(8 x-2 y=5\) (c) \(7 x-4 y=1\) (d) \(4 x-2 y=1\)

7 step solution

Problem 151

A hyperbola having the transverse axis of length \(\sqrt{2}\) has the same foci as that of the ellipse \(3 x^{2}+4 y^{2}=12\), then this hyperbola does not pass through which of the following points? (b) \(\left(-\sqrt{\frac{3}{2}}, 1\right)\) (c) \(\left(1,-\frac{1}{\sqrt{2}}\right)\) (d) \(\left(\sqrt{\frac{3}{2}}, \frac{1}{\sqrt{2}}\right)\)

3 step solution

Problem 152

Area (in sq. units) of the region outside \(\frac{|x|}{2}+\frac{|y|}{3}=1\) and inside the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\) is: (a) \(6(\pi-2)\) (b) \(3(\pi-2)\) (c) \(3(4-\pi)\) (d) \(6(4-\pi)\)

6 step solution

Problem 153

If \(e_{1}\) and \(e_{2}\) are the eccentricities of the ellipse, \(\frac{x^{2}}{18}+\frac{y^{2}}{4}=1\) and the hyperbola, \(\frac{x^{2}}{9}-\frac{y^{2}}{4}=1\) respectively and \(\left(e_{1}, e_{2}\right)\) is a point on the ellipse, \(15 x^{2}+3 y^{2}=k\), then \(k\) is equal to (a) 16 (b) 17 (c) 15 (d) 14

4 step solution

Problem 154

The length of the minor axis (along \(y\)-axis) of an ellipse in the standard form is \(\frac{4}{\sqrt{3}}\). If this ellipse touches the line, \(x\) \(+6 y=8\); then its eccentricity is: (a) \(\frac{1}{2} \sqrt{\frac{11}{3}}\) (b) \(\sqrt{\frac{5}{6}}\) (c) \(\frac{1}{2} \sqrt{\frac{5}{3}}\) (d) \(\frac{1}{3} \sqrt{\frac{11}{3}}\)

5 step solution

Problem 155

Let the line \(y=m x\) and the ellipse \(2 x^{2}+y^{2}=1\) intersect at a point \(P\) in the first quadrant. If the normal to this ellipse at \(P\) meets the co- ordinate axes at \(\left(-\frac{1}{3 \sqrt{2}}, 0\right)\) and \((0, \beta)\), then \(\beta\) is equal to: (a) \(\frac{2 \sqrt{2}}{3}\) (b) \(\frac{2}{\sqrt{3}}\) (c) \(\frac{2}{3}\) (d) \(\frac{\sqrt{2}}{3}\)

5 step solution

Problem 156

If the distance between the foci of an ellipse is 6 and the distance between its directrices is 12 , then the length of its latus rectum is: (a) \(\sqrt{3}\) (b) \(3 \sqrt{2}\) (c) \(\frac{3}{\sqrt{2}}\) (d) \(2 \sqrt{3}\)

7 step solution

Problem 157

If \(3 x+4 y=12 \sqrt{2}\) is \(a\) tangent to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{9}=1\) for some \(a \in R\), then the distance between the foci of the ellipse is: (a) \(2 \sqrt{7}\) (b) 4 (c) \(2 \sqrt{5}\) (d) \(2 \sqrt{2}\)

6 step solution

Problem 158

If the normal to the ellipse \(3 x^{2}+4 y^{2}=12\) at a point P on it is parallel to the line, \(2 x+y=4\) and the tangent to the ellipse at P passes through \(\mathrm{Q}(4,4)\) then \(\mathrm{PQ}\) is equal to: (a) \(\frac{5 \sqrt{5}}{2}\) (b) \(\frac{\sqrt{61}}{2}\) (c) \(\frac{\sqrt{221}}{2}\) (d) \(\frac{\sqrt{157}}{2}\)

6 step solution

Problem 159

An ellipse, with foci at \((0,2)\) and \((0,-2)\) and minor axis of length 4 , passes through which of the following points ? (a) \((\sqrt{2}, 2)\) (b) \((2, \sqrt{2})\) (c) \((2,2 \sqrt{2})\) (d) \((1,2 \sqrt{2})\)

4 step solution

Problem 160

If the line \(x-2 y=12\) is tangent to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) at the point \(\left(3, \frac{-9}{2}\right)\), then the length of the latus rectum of the ellipse is : (a) 9 (b) \(12 \sqrt{2}\) (c) 5 (d) \(8 \sqrt{3}\)

5 step solution

Problem 161

The tangent and normal to the ellipse \(3 x^{2}+5 y^{2}=32\) at the point \(\mathrm{P}(2,2)\) meet the \(\mathrm{x}\)-axis at \(\mathrm{Q}\) and \(\mathrm{R}\), respectively. Then the area (in sq. units) of the triangle \(\mathrm{PQR}\) is: (a) \(\frac{34}{15}\) (b) \(\frac{14}{3}\) (c) \(\frac{16}{3}\) (d) \(\frac{68}{15}\)

6 step solution

Problem 162

If the tangent to the parabola \(y^{2}=x\) at a point \((\alpha, \beta),(\beta\rangle\) 0 ) is also a tangent to the ellipse, \(x^{2}+2 y^{2}=1\), then \(\alpha\) is equal to: [April 09, 2019 (II)] (a) \(\sqrt{2}-1\) (b) \(2 \sqrt{2}-1\) (c) \(2 \sqrt{2}+1\) (d) \(\sqrt{2}+1\)

5 step solution

Problem 163

In an ellipse, with centre at the origin, if the difference of the lengths of major axis and minor axis is 10 and one of the foci is at \((0,5 \sqrt{3})\), then the length of its latus rectum is: \(\quad\) [April 08,2019 (II)] (a) 10 (b) 5 (c) 8 (d) 6

5 step solution

Problem 164

Let \(\mathrm{S}\) and \(\mathrm{S}^{\prime}\) be the foci of an ellipse and \(\mathrm{B}\) be any one of the extremities of its minor axis. If \(\Delta \mathrm{S}^{\prime} \mathrm{BS}\) is a right angled triangle with right angle at \(\mathrm{B}\) and area \(\left(\triangle \mathrm{S}^{\prime} \mathrm{BS}\right)=8 \mathrm{sq}\). units, hen the length of a latus rectum of the ellipse is : (a) 4 (b) \(2 \sqrt{2}\) (c) \(4 \sqrt{2}\) (d) 2

6 step solution

Problem 166

Two sets \(\mathrm{A}\) and \(\mathrm{B}\) are as under : \(\mathrm{A}=\\{(\mathrm{a}, \mathrm{b}) \in \mathrm{R} \times \mathrm{R}:|\mathrm{a}-5|<1\) and \(|\mathrm{b}-5|<1\\}\) \(\mathrm{B}=\left\\{(\mathrm{a}, \mathrm{b}) \in \mathrm{R} \times \mathrm{R}: 4(\mathrm{a}-6)^{2}+9(\mathrm{~b}-5)^{2} \leq 36\right\\}\). Then : (a) \(\mathrm{A} \subset \mathrm{B}\) (b) \(\mathrm{A} \cap \mathrm{B}=\phi\) (an empty set) (c) neither \(\mathrm{A} \subset \mathrm{B}\) nor \(\mathrm{B} \subset \mathrm{A}\) (d) \(\mathrm{B} \subset \mathrm{A}\)

3 step solution

Problem 167

If the length of the latus rectum of an ellipse is 4 units and the distance between a focus and its nearest vertex on the major axis is \(\frac{3}{2}\) units, then its eccentricity is? (a) \(\frac{1}{2}\) (b) \(\frac{2}{3}\) (c) \(\frac{1}{9}\) (d) \(\frac{1}{3}\)

6 step solution

Problem 168

The eccentricity of an ellipse having centre at the origin, axes along the co- ordinate axes and passing through the points \((4,-1)\) and \((-2,2)\) is: (a) \(\frac{1}{2}\) (b) \(\frac{2}{\sqrt{5}}\) (c) \(\frac{\sqrt{3}}{2}\) (d) \(\frac{\sqrt{3}}{4}\)

5 step solution

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