Chapter 20

A Complete Resource Book in Mathematics for JEE Main · 90 exercises

Problem 62

The orbit of the earth is an ellipse with eccentricity \(\frac{1}{60}\) with the sum at one focus, the major axis being approximately \(186 \times 10^{6}\) miles in length. The shortest and longest distance of the earth from the sun is (A) \(9145 \times 10^{4}\) miles, \(9455 \times 10^{4}\) miles (B) \(9147 \times 10^{4}\) miles, \(9457 \times 10^{4}\) miles (C) \(9145 \times 10^{6}\) miles, \(9455 \times 10^{6}\) miles (D) none of these

6 step solution

Problem 63

\(\mathrm{PN}\) is the ordinate of any point \(P\) on the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) and \(A A^{\prime}\) is its transverse axis. If \(Q\) divides \(A P\) in the ratio \(a^{2}: b^{2}\), then \(N Q\) is (A) \(\perp\) to \(A^{\prime} P\) (B) parallel to \(A^{\prime} P\) (C) \(\perp\) to \(O P\) (D) none of these

7 step solution

Problem 64

An ellipse has eccentricity \(\frac{1}{2}\) and one focus at the point \(P\left(\frac{1}{2}, 1\right) .\) Its one directrix is the common tan gent, nearer to the point \(P\), to the circle \(x^{2}+y^{2}=1\) and the hyperbola \(x^{2}-y^{2}=1\). The equation of the ellipse in the standard form is (A) \(\frac{\left(x-\frac{1}{3}\right)^{2}}{\frac{1}{9}}+\frac{(y-1)^{2}}{\frac{1}{12}}=1\) (B) \(\frac{\left(x-\frac{1}{3}\right)^{2}}{\frac{1}{12}}+\frac{(y-1)^{2}}{\frac{1}{9}}=1\) (C) \(\frac{(x-1)^{2}}{\frac{1}{9}}+\frac{\left(y-\frac{1}{3}\right)^{2}}{\frac{1}{12}}=1\) (D) none of these

5 step solution

Problem 65

A variable straight line of slope 4 intersects the hyperbola \(x y=1\) at two points. The locus of the point which divides the line segment between these two points in the ratio \(1: 2\) is (A) \(16 x^{2}+10 x y+y^{2}=2\) (B) \(16 x^{2}-10 x y+y^{2}=2\) (C) \(16 x^{2}+10 x y+y^{2}=4\) (D) none of these

7 step solution

Problem 66

If the parabola \(x^{2}=a y\) makes an intercept of length \(\sqrt{40}\) on the line \(y-2 x=1\), then \(a\) is equal to (A) 1 (B) \(-2\) (C) \(-1\) (D) 2

6 step solution

Problem 67

The asymptotes of the hyperbola \(x y-3 x+4 y+2=0\) are (A) \(x=-4\) (B) \(x=4\) (C) \(y=-3\) (D) \(y=3\)

5 step solution

Problem 68

If the line \(a x+b y+c=0\) is a normal to the hyperbola \(x y=1\), then (A) \(a>0, b<0\) (B) \(a>0, b>0\) (C) \(a<0, b<0\) (D) \(a<0, b>0\)

6 step solution

Problem 69

Consider a circle with its centre lying on the focus of the parabola \(y^{2}=2 p x\) such that it touches the directrix of the parabola. Then, a point of intersection of the circle and the parabola is (A) \(\left(\frac{p}{2}, p\right)\) (B) \(\left(\frac{p}{2},-p\right)\) (C) \(\left(-\frac{p}{2}, p\right)\) (D) \(\left(-\frac{p}{2},-\frac{p}{2}\right)\)

8 step solution

Problem 71

Let \(R(h, k)\) be the middle point of the chord \(P Q\) of the parabola \(y^{2}=4 a x\). Equation of \(P Q\) is $$ (y-k)=m(x-h) $$ where \(m\) is the slope of \(P Q\) \(\because R\) lies on the diameter \(y=\frac{2 a}{m}\) bisecting \(P Q\), $$ \therefore k=\frac{2 a}{m} \Leftrightarrow m=\frac{2 a}{k} $$ Subsituting this value of \(m\) in (1), we have \(y-k=\frac{2 a}{k}(x-h)\) or, \(k(y-k)=2 a(x-h)\) or \(k y-2 a x+\left(2 a h-k^{2}\right)=0\) which is the required equation. The locus of the middle point of chords of the parabola which subtend a constant angle \(\alpha\) at the vertex is \(\left(y^{2}-2 a x+\right.\) \(8 a^{2}\) ) \(^{2} \tan ^{2} \alpha=k a^{2}\left(4 a x-y^{2}\right)\), where \(k=\) (A) 4 (B) 8 (C) 16 (D) none of these

6 step solution

Problem 72

Let \(R(h, k)\) be the middle point of the chord \(P Q\) of the parabola \(y^{2}=4 a x\). Equation of \(P Q\) is $$ (y-k)=m(x-h) $$ where \(m\) is the slope of \(P Q\) \(\because R\) lies on the diameter \(y=\frac{2 a}{m}\) bisecting \(P Q\), $$ \therefore k=\frac{2 a}{m} \Leftrightarrow m=\frac{2 a}{k} $$ Subsituting this value of \(m\) in (1), we have \(y-k=\frac{2 a}{k}(x-h)\) or, \(k(y-k)=2 a(x-h)\) or \(k y-2 a x+\left(2 a h-k^{2}\right)=0\) which is the required equation. The locus of the middle point of chords of the parabola which passes through the focus is (A) \(y^{2}=a(x-a)\) (B) \(y^{2}=2 a(x-a)\) (C) \(y^{2}=4 a(x-a)\) (D) none of these

7 step solution

Problem 73

Let \(R(h, k)\) be the middle point of the chord \(P Q\) of the parabola \(y^{2}=4 a x\). Equation of \(P Q\) is $$ (y-k)=m(x-h) $$ where \(m\) is the slope of \(P Q\) \(\because R\) lies on the diameter \(y=\frac{2 a}{m}\) bisecting \(P Q\), $$ \therefore k=\frac{2 a}{m} \Leftrightarrow m=\frac{2 a}{k} $$ Subsituting this value of \(m\) in (1), we have \(y-k=\frac{2 a}{k}(x-h)\) or, \(k(y-k)=2 a(x-h)\) or \(k y-2 a x+\left(2 a h-k^{2}\right)=0\) which is the required equation. The locus of the middle point of chords of the parabola which are such that the focal distances of their extremities are in the ratio \(2: 1\), is \(9\left(y^{2}-2 a x\right)^{2}=k a^{2}(2 x-a)(4 x+a)\), where \(k=\) (A) 4 (B) 8 (C) 16 (D) none of these

6 step solution

Problem 75

Let \(R(h, k)\) be the middle point of one of the chords, say \(P Q\) of the system of parallel chords of the ellipse Let \(m\) be the slope of these parallel chords. The locus of the middle point \((h, k)\) is \(y=-\frac{b^{2}}{a^{2} m} x\), which is called a diameter of the ellipse. Two diameters are said to be conjugate if each bisects all chords parallel to the other. The condition that two diameters \(y=m_{1} x\) and \(y=m_{2} x\) may be conjugate with respect to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is \(m_{1} m_{2}=-\frac{b^{2}}{a^{2}}\). Note that the eccentric angles of the extremities of two conjugate semi-diameters differ by a right angle. If \(C P\) and \(C D\) be any two conjugate semi-diameters of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), then the tangents at \(P\) and \(D\) intersect on the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=k\), where \(k=\) (A) 1 (B) 2 (C) 4 (D) 16

5 step solution

Problem 77

In the following questions an Assertion \((A)\) is given followed by a Reason (R). Mark your responses from the following options: (A) Assertion(A) is True and Reason(R) is True; Reason( \(\mathrm{R}\) ) is a correct explanation for Assertion(A) (B) Assertion(A) is True, Reason(R) is True; Reason(R) is not a correct explanation for Assertion(A) (C) Assertion(A) is True, Reason(R) is False (D) Assertion(A) is False, Reason(R) is True Assertion: The combined equation of the asymptotes of the hyperbola \(2 x^{2}+5 x y+2 y^{2}+4 x+5 y+2=0\) Reason: The equation of a hyperbola and its asymptotes differ in constant terms only.

4 step solution

Problem 78

In the following questions an Assertion \((A)\) is given followed by a Reason (R). Mark your responses from the following options: (A) Assertion(A) is True and Reason(R) is True; Reason( \(\mathrm{R}\) ) is a correct explanation for Assertion(A) (B) Assertion(A) is True, Reason(R) is True; Reason(R) is not a correct explanation for Assertion(A) (C) Assertion(A) is True, Reason(R) is False (D) Assertion(A) is False, Reason(R) is True Assertion:The locus of the centre of the circle described on any focal chord of a parabola \(y^{2}=4 a x\) as diameter is \(y^{2}\) \(=2 a(x-a)\) Reason: If \(A\left(a t_{1}^{2}, 2 a t_{1}\right)\) and \(B\left(a t_{2}^{2}, 2 a t_{2}\right)\) be the extremities of a focal chord for the parabola \(y^{2}=4 a x\), then \(t_{12}=t_{2}=-1\)

5 step solution

Problem 79

In the following questions an Assertion \((A)\) is given followed by a Reason (R). Mark your responses from the following options: (A) Assertion(A) is True and Reason(R) is True; Reason( \(\mathrm{R}\) ) is a correct explanation for Assertion(A) (B) Assertion(A) is True, Reason(R) is True; Reason(R) is not a correct explanation for Assertion(A) (C) Assertion(A) is True, Reason(R) is False (D) Assertion(A) is False, Reason(R) is True Assertion: The angle of intersection between the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) and the circle \(x^{2}+y^{2}=a b\) is \(\tan ^{-1} \frac{(b-a)}{\sqrt{a b}}\) Reason: The point of intersection of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) and the circle \(x^{2}+y^{2}=a b\) is \(\left(\sqrt{\frac{a^{2} b}{a+b}},\right.\), \(\left.\sqrt{\frac{a b^{2}}{a+b}}\right)\)

4 step solution

Problem 80

The radius of the circle passing through the foci of the ellipse \(\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\) and having its centre at \((0,3)\), is: (A) 4 unit (B) 3 unit (C) \(\sqrt{12}\) unit (D) \(\frac{7}{2}\) unit

5 step solution

Problem 81

The equation of the ellipse whose foci are \((\pm 2,0)\) and eccentricity is \(\frac{1}{2}\) is: (A) \(\frac{x^{2}}{12}+\frac{y^{2}}{16}=1\) (B) \(\frac{x^{2}}{16}+\frac{y^{2}}{12}=1\) (C) \(\frac{x^{2}}{16}+\frac{y^{2}}{8}=1\) (D) none of these

6 step solution

Problem 82

The equation of the chord joining two points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) on the rectangular hyperbola \(x y=c^{2}\) is : (A) \(\frac{x}{x_{1}+x_{2}}+\frac{y}{y_{1}+y_{2}}=1\) (B) \(\frac{x}{x_{1}-x_{2}}+\frac{y}{y_{1}-y_{2}}=1\) (C) \(\frac{x}{y_{1}+y_{2}}+\frac{y}{x_{1}+x_{2}}=1\) (D) \(\frac{x}{y_{i}-y_{2}}+\frac{y}{x_{1}-x_{2}}=1\)

5 step solution

Problem 84

The foci of the ellipse \(\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1\) and the hyperbola \(\frac{x^{2}}{144}-\frac{y^{2}}{81}=\frac{1}{25}\) coincide. Then the value of \(b^{2}\) is (A) 1 (B) 5 (C) 7 (D) 9

4 step solution

Problem 85

A point on the parabola \(y^{2}=18 x\) at which the ordinate increases at twice the rate of the abscissa is (A) \((2,4)\) (B) \((2,-4)\) (C) \(\left(\frac{-9}{8}, \frac{9}{2}\right)\) (D) \(\left(\frac{9}{8}, \frac{9}{2}\right)\)

6 step solution

Problem 86

If \(a \neq 0\) and the line \(2 b x+3 c y+4 d=0\) passes through the points of intersection of the parabolas \(y^{2}=4 a x\) and \(x^{2}=4 a y\), then (A) \(d^{2}+(2 b+3 c)^{2}=0\) (B) \(d^{f}+(3 b+2 c)^{2}=0\) (C) \(d^{2}+(2 b-3 c)^{2}=0\) (D) \(f+(3 b-2 c)^{2}=0\)

4 step solution

Problem 88

Area of the greatest rectangle that can be inscribed in the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is (A) \(2 a b\) (B) \(a b\) (C) \(\sqrt{a b}\) (D) \(\frac{a}{b}\)

8 step solution

Problem 89

The locus of a point \(P(\alpha, \beta)\) moving under the condition that the line \(y=\alpha x+\beta\) is a tangent to the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) is (A) an ellipse (B) a circle (C) a parabola (D) a hyperbola

4 step solution

Problem 90

An ellipse has \(O B\) as semi minor axis, \(F\) and \(F^{\prime}\) its focii and the angle \(F B F^{\prime}\) is a right angle. Then the eccentricity of the ellipse is (A) \(\frac{1}{\sqrt{2}}\) (B) \(\frac{1}{2}\) (C) \(\frac{1}{4}\) (D) \(\frac{1}{\sqrt{3}}\)

5 step solution

Problem 91

In an ellipse, the distance between its foci is 6 and minor axis is 8 . Then its eccentricity is (A) \(\frac{3}{5}\) (B) \(\frac{1}{2}\) (C) \(\frac{4}{5}\) (D) \(\frac{1}{\sqrt{5}}\)

6 step solution

Problem 92

For the hyperbola \(\frac{x^{2}}{\cos ^{2} \alpha}-\frac{y^{2}}{\sin ^{2} \alpha}=1\), which of the following remains constant when \(\alpha\). varies? (A) eccentricity (B) directrix (C) abscissae of vertices (D) abscissae of foci

5 step solution

Problem 93

A focus of an ellipse is at the origin. The directrix is the line \(x=4\) and the eccentricity is \(\frac{1}{2}\), Then the length of the semi-major axis is (A) \(\frac{8}{3}\) (B) \(\frac{2}{3}\) (C) \(\frac{4}{3}\) (D) \(\frac{5}{3}\)

5 step solution

Problem 94

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) \((2,0)\)

4 step solution

Problem 95

The ellipse \(x^{2}+4 y^{2}=4\) is inscribed in a rectangle aligned with the coordinate axes, which in turn in inscribed in another ellipse that passes through the point \((4,0)\). Then the equation of the ellipse is (A) \(x^{2}+16 y^{2}=16\) (B) \(x^{2}+12 y^{2}=16\) (C) \(4 x^{2}+48 y^{2}=48\) (D) \(4 x^{2}+64 y^{2}=48\)

5 step solution

Problem 96

If two tangents drawn from a point \(P\) to the parabola \(y_{1}^{2}\) \(=4 x\) are at right angles, then the locus of the point \(P\) is (A) \(2 x+1=0\) (B) \(x=-1\) (C) \(2 x-1=0\) (D) \(x=1\)

5 step solution

Problem 98

An ellipse is drawn by considering a diameter of the circle \((x-1)^{2}+y^{2}=1\) as its semi-minor axis and a diameter of the circle \(x^{2}+(y-2)^{2}=4\) as its semi-major axis. If the centre of the ellipse is the origin and its axes are the coordinate axes, then the equation of the ellipse is (A) \(4 x^{2}+y^{2}=4\) (B) \(x^{2}+4 y^{2}=8\) (C) \(4 x^{2}+y^{2}=8\) (D) \(x^{2}+4 y^{2}=16\)

6 step solution

Problem 99

Given: A circle, \(2 x^{2}+2 y^{2}=5\) and a parabola, \(y^{2}=4 \sqrt{5 x}\) Statement - I: An equation of a common tangent to these curves is \(y=x+\sqrt{5}\). Statement - II: 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 - \(I\) is True; Statement -II is true; Statement-II is not a correct explanation for Statement-I (B) Statement \(-\mathrm{I}\) is True; Statement -II is False. (C) Statement -I is False; Statement -II is True (D) Statement \(-\mathrm{I}\) is True; Statement -II is True; Statement-II is a correct explanation for Statement-I

6 step solution

Problem 100

The circle passing through the foci of the ellipse \(\frac{x^{2}}{16}+\frac{y^{2}}{4}=1\) with center at \((0,3)\) has equation (A) \(x^{2}+y^{2}-6 y+7=0\) (B) \(x^{2}+y^{2}-6 y-5=0\) (C) \(x^{2}+y^{2}-6 y+5=0\) (D) \(x^{2}+y^{2}-6 y-7=0\)

5 step solution

Problem 101

The locus of the foot of the perpendicular drawn from the centre of the ellipse \(x^{2}+3 y^{2}=6\) on any tangent to it is (A) \(\left(x^{2}-y^{2}\right)^{2}=6 x^{2}+2 y^{2}\) (B) \(\left(x^{2}-y^{2}\right)^{2}=6 x^{2}-2 y^{2}\) (C) \(\left(x^{2}+y^{2}\right)^{2}=6 x^{2}+2 y^{2}\) (D) \(\left(x^{2}+y^{2}\right)^{2}=6 x^{2}-2 y^{2}\)

5 step solution

Problem 102

The slope of the line touching both the parabolas \(y^{2}=\) \(4 x\) and \(x^{2}=-32 y\) is (A) \(\frac{1}{2}\) (B) \(\frac{3}{2}\) (C) \(\frac{1}{8}\) (D) \(\frac{2}{3}\)

7 step solution

Problem 103

Let \(O\) be the vertex and \(Q\) be any point on the parabola, \(x^{2}=8 y\). If the point \(P\) divides the line segment \(O Q\) internally in the ratio \(1: 3\), then the locus of \(P\) is: (A) \(y^{2}=x\) (B) \(y^{2}=2 x\) (C) \(x^{2}=2 y\) (D) \(x^{2}=y\)

5 step solution

Problem 104

The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latus rectum to the ellipse \(\frac{x^{2}}{9}+\frac{y^{2}}{5}=1\), is: (A) 18 (B) \(\frac{27}{2}\) (C) 27 (D) \(\frac{27}{4}\)

6 step solution

Problem 105

The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is (A) \(\sqrt{3}\) (B) \(\frac{4}{3}\) (C) \(\frac{4}{\sqrt{3}}\) (D) \(\frac{2}{\sqrt{3}}\)

6 step solution

Problem 106

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

8 step solution

Problem 107

The centres of those circles which touch the circle, \(x^{2}\) \(+y^{2}-8 x-8 y-4=0\), externally and also touch the \(x\)-axis, lie on (A) A parabola (B) A circle (C) An ellipse which is not a circle (D) A hyperbola

7 step solution

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