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Ellipse

120530 The locus of the extremities of the latusrectum of the family of ellipses
\(b^2 x^2+y^2=a^2 b^2\) having a given major axis is

1 \(x^2 \pm a y=a^2\)

2 \(y^2 \pm b x=a^2\)

3 \(x^2 \pm b y=a^2\)

4 \(x^2 \pm a x=b^2\)

CG PET- 2015

Ellipse

120531 The locus of the mid point of the line segment joining the point \((4,3)\) and the points on the ellipse \(x^2+2 y^2=4\) is an ellipse with eccentricity:

1 \(\frac{\sqrt{3}}{2}\)

2 \(\frac{1}{2 \sqrt{2}}\)

3 \(\frac{1}{\sqrt{2}}\)

4 \(\frac{1}{2}\)

JEE Main-26.06.2022

Ellipse

120532 Let \(P\left(\frac{2 \sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}}\right), Q, R\) and \(S\) be four points on the ellipse \(9 x^2+4 y^2=36\), Let \(P Q\) and RS be mutually perpendicular and pass through the origin. If \(\frac{1}{(P Q)^2}+\frac{1}{(R S)^2}=\frac{p}{q}\), where \(p\) and \(q\) are coprime, then \(p+q\) is equal to

1 143

2 137

3 157

4 147

Explanation:

C Given,
The equation of ellipse is \(9 x^2+4 y^2=36\)
The coordinate of point \(\mathrm{P}\) is \(\left[\frac{2 \sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}}\right]\)
\(\because \mathrm{PQ}\) passes through origin,
\(\therefore\) Slope of \(P Q=\frac{6 / \sqrt{7}-0}{2 \sqrt{3} / \sqrt{7}-0}=\sqrt{3}\)
\(\therefore\) Equation of \(\mathrm{PQ}\) is \(\mathrm{y}=\sqrt{3} \mathrm{x}\)
Now, from equation (i)
\(9 x^2+4(\sqrt{3} x)^2=36\)
\(9 x^2+4 \times 3 x^2=36\)
\(21 x^2=36\)
\(x= \pm \frac{2 \sqrt{3}}{\sqrt{7}}\)
\(y= \pm \frac{6}{\sqrt{7}}\)
\(\therefore\) The co-ordinate of Point Q is \(\left[\frac{-2 \sqrt{3}}{\sqrt{7}}, \frac{-6}{\sqrt{7}}\right]\)
\(\therefore\) Now,
The distance \(P Q=\sqrt{\left[\frac{2 \sqrt{3}}{\sqrt{7}}+\frac{2 \sqrt{3}}{\sqrt{7}}\right]^2+\left[\frac{6}{\sqrt{7}}+\frac{6}{\sqrt{7}}\right]^2}\)
\(=\sqrt{\left(\frac{4 \sqrt{3}}{\sqrt{7}}\right)^2+\left(\frac{12}{\sqrt{7}}\right)^2}\)
\(=\sqrt{\frac{48}{7}+\frac{144}{7}}\)
\((\mathrm{PQ})^2=\frac{192}{7}\)
Now, \(\mathrm{RS} \perp \mathrm{PQ}\)
\(\therefore\) Equation of RS is \(y=\frac{-x}{\sqrt{3}}\)
Now, from equation (i)
\(9 x^2+4\left(\frac{-x}{\sqrt{3}}\right)^2=36\)
\(9 x^2+\frac{4}{3} x^2=36\)
\(31 x^2=36 \times 3\)
\(x= \pm \frac{6 \sqrt{3}}{\sqrt{31}}, y= \pm \frac{6}{\sqrt{31}}\)
\(\mathrm{R}=\left[\frac{6 \sqrt{3}}{\sqrt{31}}, \frac{-6}{\sqrt{31}}\right] \text { and } \mathrm{S}=\left[\frac{-6 \sqrt{3}}{\sqrt{31}}, \frac{6}{\sqrt{31}}\right]\)
\((\mathrm{RS})^2=\left[\frac{6 \sqrt{3}}{\sqrt{31}}+\frac{6 \sqrt{3}}{\sqrt{31}}\right]^2+\left[\frac{-6}{\sqrt{31}}-\frac{6}{\sqrt{31}}\right]^2\)
\(=\frac{576}{31}\)
\(\therefore \frac{1}{(\mathrm{PQ})^2}+\frac{1}{(\mathrm{RS})^2}\)
\(=\left(\frac{7}{192}\right)^2+\left(\frac{31}{576}\right)^2\)
\(=\frac{1}{(192)^2}\left[49+\frac{31^2}{9}\right]=\frac{13}{144}=\frac{\mathrm{p}}{\mathrm{q}}\)
As given \(\mathrm{P}\) and \(\mathrm{Q}\) are comprime
\(\therefore \mathrm{P}+\mathrm{q}=13+144=157\)

JEE Main-12.04.2023

Ellipse

120533 let \(P Q\) be a focal chord of the parabola \(y^2=4 x\) such that it subtends an angle of \(\frac{\pi}{2}\) at the point \((3,0)\). Let the line segment \(P Q\) be also a focal chord of the ellipse \(E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a^2>b^2\). If \(e\) is the eccentricity of the ellipse \(E\), then the value of \(\frac{1}{e^2}\) is equal to :

1 \(1+\sqrt{2}\)

2 \(3+2 \sqrt{2}\)

3 \(1+2 \sqrt{3}\)

4 \(4+5 \sqrt{3}\)

JEE Main-28.06.2022

Ellipse

120530 The locus of the extremities of the latusrectum of the family of ellipses
\(b^2 x^2+y^2=a^2 b^2\) having a given major axis is

1 \(x^2 \pm a y=a^2\)

2 \(y^2 \pm b x=a^2\)

3 \(x^2 \pm b y=a^2\)

4 \(x^2 \pm a x=b^2\)

CG PET- 2015

Ellipse

120531 The locus of the mid point of the line segment joining the point \((4,3)\) and the points on the ellipse \(x^2+2 y^2=4\) is an ellipse with eccentricity:

1 \(\frac{\sqrt{3}}{2}\)

2 \(\frac{1}{2 \sqrt{2}}\)

3 \(\frac{1}{\sqrt{2}}\)

4 \(\frac{1}{2}\)

JEE Main-26.06.2022

Ellipse

120532 Let \(P\left(\frac{2 \sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}}\right), Q, R\) and \(S\) be four points on the ellipse \(9 x^2+4 y^2=36\), Let \(P Q\) and RS be mutually perpendicular and pass through the origin. If \(\frac{1}{(P Q)^2}+\frac{1}{(R S)^2}=\frac{p}{q}\), where \(p\) and \(q\) are coprime, then \(p+q\) is equal to

1 143

2 137

3 157

4 147

Explanation:

C Given,
The equation of ellipse is \(9 x^2+4 y^2=36\)
The coordinate of point \(\mathrm{P}\) is \(\left[\frac{2 \sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}}\right]\)
\(\because \mathrm{PQ}\) passes through origin,
\(\therefore\) Slope of \(P Q=\frac{6 / \sqrt{7}-0}{2 \sqrt{3} / \sqrt{7}-0}=\sqrt{3}\)
\(\therefore\) Equation of \(\mathrm{PQ}\) is \(\mathrm{y}=\sqrt{3} \mathrm{x}\)
Now, from equation (i)
\(9 x^2+4(\sqrt{3} x)^2=36\)
\(9 x^2+4 \times 3 x^2=36\)
\(21 x^2=36\)
\(x= \pm \frac{2 \sqrt{3}}{\sqrt{7}}\)
\(y= \pm \frac{6}{\sqrt{7}}\)
\(\therefore\) The co-ordinate of Point Q is \(\left[\frac{-2 \sqrt{3}}{\sqrt{7}}, \frac{-6}{\sqrt{7}}\right]\)
\(\therefore\) Now,
The distance \(P Q=\sqrt{\left[\frac{2 \sqrt{3}}{\sqrt{7}}+\frac{2 \sqrt{3}}{\sqrt{7}}\right]^2+\left[\frac{6}{\sqrt{7}}+\frac{6}{\sqrt{7}}\right]^2}\)
\(=\sqrt{\left(\frac{4 \sqrt{3}}{\sqrt{7}}\right)^2+\left(\frac{12}{\sqrt{7}}\right)^2}\)
\(=\sqrt{\frac{48}{7}+\frac{144}{7}}\)
\((\mathrm{PQ})^2=\frac{192}{7}\)
Now, \(\mathrm{RS} \perp \mathrm{PQ}\)
\(\therefore\) Equation of RS is \(y=\frac{-x}{\sqrt{3}}\)
Now, from equation (i)
\(9 x^2+4\left(\frac{-x}{\sqrt{3}}\right)^2=36\)
\(9 x^2+\frac{4}{3} x^2=36\)
\(31 x^2=36 \times 3\)
\(x= \pm \frac{6 \sqrt{3}}{\sqrt{31}}, y= \pm \frac{6}{\sqrt{31}}\)
\(\mathrm{R}=\left[\frac{6 \sqrt{3}}{\sqrt{31}}, \frac{-6}{\sqrt{31}}\right] \text { and } \mathrm{S}=\left[\frac{-6 \sqrt{3}}{\sqrt{31}}, \frac{6}{\sqrt{31}}\right]\)
\((\mathrm{RS})^2=\left[\frac{6 \sqrt{3}}{\sqrt{31}}+\frac{6 \sqrt{3}}{\sqrt{31}}\right]^2+\left[\frac{-6}{\sqrt{31}}-\frac{6}{\sqrt{31}}\right]^2\)
\(=\frac{576}{31}\)
\(\therefore \frac{1}{(\mathrm{PQ})^2}+\frac{1}{(\mathrm{RS})^2}\)
\(=\left(\frac{7}{192}\right)^2+\left(\frac{31}{576}\right)^2\)
\(=\frac{1}{(192)^2}\left[49+\frac{31^2}{9}\right]=\frac{13}{144}=\frac{\mathrm{p}}{\mathrm{q}}\)
As given \(\mathrm{P}\) and \(\mathrm{Q}\) are comprime
\(\therefore \mathrm{P}+\mathrm{q}=13+144=157\)

JEE Main-12.04.2023

Ellipse

120533 let \(P Q\) be a focal chord of the parabola \(y^2=4 x\) such that it subtends an angle of \(\frac{\pi}{2}\) at the point \((3,0)\). Let the line segment \(P Q\) be also a focal chord of the ellipse \(E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a^2>b^2\). If \(e\) is the eccentricity of the ellipse \(E\), then the value of \(\frac{1}{e^2}\) is equal to :

1 \(1+\sqrt{2}\)

2 \(3+2 \sqrt{2}\)

3 \(1+2 \sqrt{3}\)

4 \(4+5 \sqrt{3}\)

JEE Main-28.06.2022

Ellipse

120530 The locus of the extremities of the latusrectum of the family of ellipses
\(b^2 x^2+y^2=a^2 b^2\) having a given major axis is

1 \(x^2 \pm a y=a^2\)

2 \(y^2 \pm b x=a^2\)

3 \(x^2 \pm b y=a^2\)

4 \(x^2 \pm a x=b^2\)

CG PET- 2015

Ellipse

120531 The locus of the mid point of the line segment joining the point \((4,3)\) and the points on the ellipse \(x^2+2 y^2=4\) is an ellipse with eccentricity:

1 \(\frac{\sqrt{3}}{2}\)

2 \(\frac{1}{2 \sqrt{2}}\)

3 \(\frac{1}{\sqrt{2}}\)

4 \(\frac{1}{2}\)

JEE Main-26.06.2022

Ellipse

120532 Let \(P\left(\frac{2 \sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}}\right), Q, R\) and \(S\) be four points on the ellipse \(9 x^2+4 y^2=36\), Let \(P Q\) and RS be mutually perpendicular and pass through the origin. If \(\frac{1}{(P Q)^2}+\frac{1}{(R S)^2}=\frac{p}{q}\), where \(p\) and \(q\) are coprime, then \(p+q\) is equal to

1 143

2 137

3 157

4 147

Explanation:

C Given,
The equation of ellipse is \(9 x^2+4 y^2=36\)
The coordinate of point \(\mathrm{P}\) is \(\left[\frac{2 \sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}}\right]\)
\(\because \mathrm{PQ}\) passes through origin,
\(\therefore\) Slope of \(P Q=\frac{6 / \sqrt{7}-0}{2 \sqrt{3} / \sqrt{7}-0}=\sqrt{3}\)
\(\therefore\) Equation of \(\mathrm{PQ}\) is \(\mathrm{y}=\sqrt{3} \mathrm{x}\)
Now, from equation (i)
\(9 x^2+4(\sqrt{3} x)^2=36\)
\(9 x^2+4 \times 3 x^2=36\)
\(21 x^2=36\)
\(x= \pm \frac{2 \sqrt{3}}{\sqrt{7}}\)
\(y= \pm \frac{6}{\sqrt{7}}\)
\(\therefore\) The co-ordinate of Point Q is \(\left[\frac{-2 \sqrt{3}}{\sqrt{7}}, \frac{-6}{\sqrt{7}}\right]\)
\(\therefore\) Now,
The distance \(P Q=\sqrt{\left[\frac{2 \sqrt{3}}{\sqrt{7}}+\frac{2 \sqrt{3}}{\sqrt{7}}\right]^2+\left[\frac{6}{\sqrt{7}}+\frac{6}{\sqrt{7}}\right]^2}\)
\(=\sqrt{\left(\frac{4 \sqrt{3}}{\sqrt{7}}\right)^2+\left(\frac{12}{\sqrt{7}}\right)^2}\)
\(=\sqrt{\frac{48}{7}+\frac{144}{7}}\)
\((\mathrm{PQ})^2=\frac{192}{7}\)
Now, \(\mathrm{RS} \perp \mathrm{PQ}\)
\(\therefore\) Equation of RS is \(y=\frac{-x}{\sqrt{3}}\)
Now, from equation (i)
\(9 x^2+4\left(\frac{-x}{\sqrt{3}}\right)^2=36\)
\(9 x^2+\frac{4}{3} x^2=36\)
\(31 x^2=36 \times 3\)
\(x= \pm \frac{6 \sqrt{3}}{\sqrt{31}}, y= \pm \frac{6}{\sqrt{31}}\)
\(\mathrm{R}=\left[\frac{6 \sqrt{3}}{\sqrt{31}}, \frac{-6}{\sqrt{31}}\right] \text { and } \mathrm{S}=\left[\frac{-6 \sqrt{3}}{\sqrt{31}}, \frac{6}{\sqrt{31}}\right]\)
\((\mathrm{RS})^2=\left[\frac{6 \sqrt{3}}{\sqrt{31}}+\frac{6 \sqrt{3}}{\sqrt{31}}\right]^2+\left[\frac{-6}{\sqrt{31}}-\frac{6}{\sqrt{31}}\right]^2\)
\(=\frac{576}{31}\)
\(\therefore \frac{1}{(\mathrm{PQ})^2}+\frac{1}{(\mathrm{RS})^2}\)
\(=\left(\frac{7}{192}\right)^2+\left(\frac{31}{576}\right)^2\)
\(=\frac{1}{(192)^2}\left[49+\frac{31^2}{9}\right]=\frac{13}{144}=\frac{\mathrm{p}}{\mathrm{q}}\)
As given \(\mathrm{P}\) and \(\mathrm{Q}\) are comprime
\(\therefore \mathrm{P}+\mathrm{q}=13+144=157\)

JEE Main-12.04.2023

Ellipse

120533 let \(P Q\) be a focal chord of the parabola \(y^2=4 x\) such that it subtends an angle of \(\frac{\pi}{2}\) at the point \((3,0)\). Let the line segment \(P Q\) be also a focal chord of the ellipse \(E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a^2>b^2\). If \(e\) is the eccentricity of the ellipse \(E\), then the value of \(\frac{1}{e^2}\) is equal to :

1 \(1+\sqrt{2}\)

2 \(3+2 \sqrt{2}\)

3 \(1+2 \sqrt{3}\)

4 \(4+5 \sqrt{3}\)

JEE Main-28.06.2022

Ellipse

120530 The locus of the extremities of the latusrectum of the family of ellipses
\(b^2 x^2+y^2=a^2 b^2\) having a given major axis is

1 \(x^2 \pm a y=a^2\)

2 \(y^2 \pm b x=a^2\)

3 \(x^2 \pm b y=a^2\)

4 \(x^2 \pm a x=b^2\)

CG PET- 2015

Ellipse

120531 The locus of the mid point of the line segment joining the point \((4,3)\) and the points on the ellipse \(x^2+2 y^2=4\) is an ellipse with eccentricity:

1 \(\frac{\sqrt{3}}{2}\)

2 \(\frac{1}{2 \sqrt{2}}\)

3 \(\frac{1}{\sqrt{2}}\)

4 \(\frac{1}{2}\)

JEE Main-26.06.2022

Ellipse

120532 Let \(P\left(\frac{2 \sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}}\right), Q, R\) and \(S\) be four points on the ellipse \(9 x^2+4 y^2=36\), Let \(P Q\) and RS be mutually perpendicular and pass through the origin. If \(\frac{1}{(P Q)^2}+\frac{1}{(R S)^2}=\frac{p}{q}\), where \(p\) and \(q\) are coprime, then \(p+q\) is equal to

1 143

2 137

3 157

4 147

Explanation:

C Given,
The equation of ellipse is \(9 x^2+4 y^2=36\)
The coordinate of point \(\mathrm{P}\) is \(\left[\frac{2 \sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}}\right]\)
\(\because \mathrm{PQ}\) passes through origin,
\(\therefore\) Slope of \(P Q=\frac{6 / \sqrt{7}-0}{2 \sqrt{3} / \sqrt{7}-0}=\sqrt{3}\)
\(\therefore\) Equation of \(\mathrm{PQ}\) is \(\mathrm{y}=\sqrt{3} \mathrm{x}\)
Now, from equation (i)
\(9 x^2+4(\sqrt{3} x)^2=36\)
\(9 x^2+4 \times 3 x^2=36\)
\(21 x^2=36\)
\(x= \pm \frac{2 \sqrt{3}}{\sqrt{7}}\)
\(y= \pm \frac{6}{\sqrt{7}}\)
\(\therefore\) The co-ordinate of Point Q is \(\left[\frac{-2 \sqrt{3}}{\sqrt{7}}, \frac{-6}{\sqrt{7}}\right]\)
\(\therefore\) Now,
The distance \(P Q=\sqrt{\left[\frac{2 \sqrt{3}}{\sqrt{7}}+\frac{2 \sqrt{3}}{\sqrt{7}}\right]^2+\left[\frac{6}{\sqrt{7}}+\frac{6}{\sqrt{7}}\right]^2}\)
\(=\sqrt{\left(\frac{4 \sqrt{3}}{\sqrt{7}}\right)^2+\left(\frac{12}{\sqrt{7}}\right)^2}\)
\(=\sqrt{\frac{48}{7}+\frac{144}{7}}\)
\((\mathrm{PQ})^2=\frac{192}{7}\)
Now, \(\mathrm{RS} \perp \mathrm{PQ}\)
\(\therefore\) Equation of RS is \(y=\frac{-x}{\sqrt{3}}\)
Now, from equation (i)
\(9 x^2+4\left(\frac{-x}{\sqrt{3}}\right)^2=36\)
\(9 x^2+\frac{4}{3} x^2=36\)
\(31 x^2=36 \times 3\)
\(x= \pm \frac{6 \sqrt{3}}{\sqrt{31}}, y= \pm \frac{6}{\sqrt{31}}\)
\(\mathrm{R}=\left[\frac{6 \sqrt{3}}{\sqrt{31}}, \frac{-6}{\sqrt{31}}\right] \text { and } \mathrm{S}=\left[\frac{-6 \sqrt{3}}{\sqrt{31}}, \frac{6}{\sqrt{31}}\right]\)
\((\mathrm{RS})^2=\left[\frac{6 \sqrt{3}}{\sqrt{31}}+\frac{6 \sqrt{3}}{\sqrt{31}}\right]^2+\left[\frac{-6}{\sqrt{31}}-\frac{6}{\sqrt{31}}\right]^2\)
\(=\frac{576}{31}\)
\(\therefore \frac{1}{(\mathrm{PQ})^2}+\frac{1}{(\mathrm{RS})^2}\)
\(=\left(\frac{7}{192}\right)^2+\left(\frac{31}{576}\right)^2\)
\(=\frac{1}{(192)^2}\left[49+\frac{31^2}{9}\right]=\frac{13}{144}=\frac{\mathrm{p}}{\mathrm{q}}\)
As given \(\mathrm{P}\) and \(\mathrm{Q}\) are comprime
\(\therefore \mathrm{P}+\mathrm{q}=13+144=157\)

JEE Main-12.04.2023

Ellipse

120533 let \(P Q\) be a focal chord of the parabola \(y^2=4 x\) such that it subtends an angle of \(\frac{\pi}{2}\) at the point \((3,0)\). Let the line segment \(P Q\) be also a focal chord of the ellipse \(E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a^2>b^2\). If \(e\) is the eccentricity of the ellipse \(E\), then the value of \(\frac{1}{e^2}\) is equal to :

1 \(1+\sqrt{2}\)

2 \(3+2 \sqrt{2}\)

3 \(1+2 \sqrt{3}\)

4 \(4+5 \sqrt{3}\)

JEE Main-28.06.2022

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