120647 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
#[Qdiff: Hard, QCat: Numerical Based, examname: WB JEE-2016], Hence, \((\mathrm{h}, Exp:(B) Standard equation of normal to the ellipse, \(\left.\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \text { at }(a \cos \theta) b \sin \theta\right) \text { is }\), \(\frac{a x}{\cos \theta}-\frac{b y}{\sin \theta}=a^2-b^2\), Let normal is passing through a given point, \mathrm{k}) \text { then } \frac{\mathrm{ah}}{\cos \theta}-\frac{\mathrm{by}}{\sin \theta}=\mathrm{a}^2-\mathrm{b}^2\), \(= 2 \mathrm{ah}\left(1+\tan ^2 \theta / 2\right) \tan \theta / 2=\mathrm{bk}\left(1-\tan ^4 \theta / 2\right)\), \(= 2\left(\mathrm{a}^2-\mathrm{b}^2\right) \tan ^2 \theta / 2\left(1-\tan ^2 \theta / 2\right)\), Which is a four degree polynomial in \(\tan \theta / 2\) therefore it may have four real roots., the number of normal's \(4\), 918. The line \(\mathbf{y}=\mathbf{n}+\lambda\) is tangent to the ellipse \(2 x^2+3 y^2=1\) Then, \(\lambda\) is,

120648 Tangents are drawn to the ellipse \(\frac{x^2}{9}+\frac{y^2}{5}=1\) at the ends of both latus rectum. The area of the quadrilateral, so formed is
#[Qdiff: Hard, QCat: Numerical Based, examname: WB JEE-2022], Where, On comparing standard ellipse equation, \(\frac{\mathrm{x}^2}{\mathrm{~A}^2}+\frac{\mathrm{y}^2}{\mathrm{~B}^2}=1\), \(A^2=36 \& B^2=4\), According to question-, \(\frac{\mathrm{A}^2}{\mathrm{~b}^2}+\frac{\mathrm{B}^2}{\mathrm{a}^2}=\frac{\left(\mathrm{A}^2-\mathrm{B}^2\right)}{\mathrm{c}^2}\), \(\frac{36}{\mathrm{~b}^2}+\frac{4}{\mathrm{a}^2}=\frac{(36-4)^2}{\mathrm{c}^2}\), \(\frac{1}{\mathrm{~b}^2}+\frac{1}{\mathrm{a}^2}=\frac{(32)^2}{2}\), \(\frac{1}{\mathrm{a}^2}+\frac{9}{\mathrm{~b}^2}=\frac{256}{\mathrm{c}^2}\), 921. \(A B\) is a variable chord of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\). If \(A B\) subtends a right angle at the origin \(\mathrm{O}\), then \(\frac{1}{\mathrm{OA}^2}+\frac{1}{\mathrm{OB}^2}\) equals to,

120649 The equation of the tangent to the ellipse \(x^2+\) \(16 \mathrm{y}^2=16\) which makes an angle \(60^{\circ}\) with the \(X\) axis is \(\qquad\)
#[Qdiff: Hard, QCat: Numerical Based, examname: AP EAMCET-22.09.2020,Shift-I],

120651 The eccentricity of an ellipse whose centre is at the origin is \(\mathbf{1 / 2}\). If one of its directrices is \(x=-\) 4 , then the equation of the normal to it at \(\left(1, \frac{3}{2}\right)\) is

120647 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
#[Qdiff: Hard, QCat: Numerical Based, examname: WB JEE-2016], Hence, \((\mathrm{h}, Exp:(B) Standard equation of normal to the ellipse, \(\left.\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \text { at }(a \cos \theta) b \sin \theta\right) \text { is }\), \(\frac{a x}{\cos \theta}-\frac{b y}{\sin \theta}=a^2-b^2\), Let normal is passing through a given point, \mathrm{k}) \text { then } \frac{\mathrm{ah}}{\cos \theta}-\frac{\mathrm{by}}{\sin \theta}=\mathrm{a}^2-\mathrm{b}^2\), \(= 2 \mathrm{ah}\left(1+\tan ^2 \theta / 2\right) \tan \theta / 2=\mathrm{bk}\left(1-\tan ^4 \theta / 2\right)\), \(= 2\left(\mathrm{a}^2-\mathrm{b}^2\right) \tan ^2 \theta / 2\left(1-\tan ^2 \theta / 2\right)\), Which is a four degree polynomial in \(\tan \theta / 2\) therefore it may have four real roots., the number of normal's \(4\), 918. The line \(\mathbf{y}=\mathbf{n}+\lambda\) is tangent to the ellipse \(2 x^2+3 y^2=1\) Then, \(\lambda\) is,

120648 Tangents are drawn to the ellipse \(\frac{x^2}{9}+\frac{y^2}{5}=1\) at the ends of both latus rectum. The area of the quadrilateral, so formed is
#[Qdiff: Hard, QCat: Numerical Based, examname: WB JEE-2022], Where, On comparing standard ellipse equation, \(\frac{\mathrm{x}^2}{\mathrm{~A}^2}+\frac{\mathrm{y}^2}{\mathrm{~B}^2}=1\), \(A^2=36 \& B^2=4\), According to question-, \(\frac{\mathrm{A}^2}{\mathrm{~b}^2}+\frac{\mathrm{B}^2}{\mathrm{a}^2}=\frac{\left(\mathrm{A}^2-\mathrm{B}^2\right)}{\mathrm{c}^2}\), \(\frac{36}{\mathrm{~b}^2}+\frac{4}{\mathrm{a}^2}=\frac{(36-4)^2}{\mathrm{c}^2}\), \(\frac{1}{\mathrm{~b}^2}+\frac{1}{\mathrm{a}^2}=\frac{(32)^2}{2}\), \(\frac{1}{\mathrm{a}^2}+\frac{9}{\mathrm{~b}^2}=\frac{256}{\mathrm{c}^2}\), 921. \(A B\) is a variable chord of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\). If \(A B\) subtends a right angle at the origin \(\mathrm{O}\), then \(\frac{1}{\mathrm{OA}^2}+\frac{1}{\mathrm{OB}^2}\) equals to,

120649 The equation of the tangent to the ellipse \(x^2+\) \(16 \mathrm{y}^2=16\) which makes an angle \(60^{\circ}\) with the \(X\) axis is \(\qquad\)
#[Qdiff: Hard, QCat: Numerical Based, examname: AP EAMCET-22.09.2020,Shift-I],

120651 The eccentricity of an ellipse whose centre is at the origin is \(\mathbf{1 / 2}\). If one of its directrices is \(x=-\) 4 , then the equation of the normal to it at \(\left(1, \frac{3}{2}\right)\) is

120647 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
#[Qdiff: Hard, QCat: Numerical Based, examname: WB JEE-2016], Hence, \((\mathrm{h}, Exp:(B) Standard equation of normal to the ellipse, \(\left.\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \text { at }(a \cos \theta) b \sin \theta\right) \text { is }\), \(\frac{a x}{\cos \theta}-\frac{b y}{\sin \theta}=a^2-b^2\), Let normal is passing through a given point, \mathrm{k}) \text { then } \frac{\mathrm{ah}}{\cos \theta}-\frac{\mathrm{by}}{\sin \theta}=\mathrm{a}^2-\mathrm{b}^2\), \(= 2 \mathrm{ah}\left(1+\tan ^2 \theta / 2\right) \tan \theta / 2=\mathrm{bk}\left(1-\tan ^4 \theta / 2\right)\), \(= 2\left(\mathrm{a}^2-\mathrm{b}^2\right) \tan ^2 \theta / 2\left(1-\tan ^2 \theta / 2\right)\), Which is a four degree polynomial in \(\tan \theta / 2\) therefore it may have four real roots., the number of normal's \(4\), 918. The line \(\mathbf{y}=\mathbf{n}+\lambda\) is tangent to the ellipse \(2 x^2+3 y^2=1\) Then, \(\lambda\) is,

120648 Tangents are drawn to the ellipse \(\frac{x^2}{9}+\frac{y^2}{5}=1\) at the ends of both latus rectum. The area of the quadrilateral, so formed is
#[Qdiff: Hard, QCat: Numerical Based, examname: WB JEE-2022], Where, On comparing standard ellipse equation, \(\frac{\mathrm{x}^2}{\mathrm{~A}^2}+\frac{\mathrm{y}^2}{\mathrm{~B}^2}=1\), \(A^2=36 \& B^2=4\), According to question-, \(\frac{\mathrm{A}^2}{\mathrm{~b}^2}+\frac{\mathrm{B}^2}{\mathrm{a}^2}=\frac{\left(\mathrm{A}^2-\mathrm{B}^2\right)}{\mathrm{c}^2}\), \(\frac{36}{\mathrm{~b}^2}+\frac{4}{\mathrm{a}^2}=\frac{(36-4)^2}{\mathrm{c}^2}\), \(\frac{1}{\mathrm{~b}^2}+\frac{1}{\mathrm{a}^2}=\frac{(32)^2}{2}\), \(\frac{1}{\mathrm{a}^2}+\frac{9}{\mathrm{~b}^2}=\frac{256}{\mathrm{c}^2}\), 921. \(A B\) is a variable chord of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\). If \(A B\) subtends a right angle at the origin \(\mathrm{O}\), then \(\frac{1}{\mathrm{OA}^2}+\frac{1}{\mathrm{OB}^2}\) equals to,

120649 The equation of the tangent to the ellipse \(x^2+\) \(16 \mathrm{y}^2=16\) which makes an angle \(60^{\circ}\) with the \(X\) axis is \(\qquad\)
#[Qdiff: Hard, QCat: Numerical Based, examname: AP EAMCET-22.09.2020,Shift-I],

120651 The eccentricity of an ellipse whose centre is at the origin is \(\mathbf{1 / 2}\). If one of its directrices is \(x=-\) 4 , then the equation of the normal to it at \(\left(1, \frac{3}{2}\right)\) is

120647 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
#[Qdiff: Hard, QCat: Numerical Based, examname: WB JEE-2016], Hence, \((\mathrm{h}, Exp:(B) Standard equation of normal to the ellipse, \(\left.\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \text { at }(a \cos \theta) b \sin \theta\right) \text { is }\), \(\frac{a x}{\cos \theta}-\frac{b y}{\sin \theta}=a^2-b^2\), Let normal is passing through a given point, \mathrm{k}) \text { then } \frac{\mathrm{ah}}{\cos \theta}-\frac{\mathrm{by}}{\sin \theta}=\mathrm{a}^2-\mathrm{b}^2\), \(= 2 \mathrm{ah}\left(1+\tan ^2 \theta / 2\right) \tan \theta / 2=\mathrm{bk}\left(1-\tan ^4 \theta / 2\right)\), \(= 2\left(\mathrm{a}^2-\mathrm{b}^2\right) \tan ^2 \theta / 2\left(1-\tan ^2 \theta / 2\right)\), Which is a four degree polynomial in \(\tan \theta / 2\) therefore it may have four real roots., the number of normal's \(4\), 918. The line \(\mathbf{y}=\mathbf{n}+\lambda\) is tangent to the ellipse \(2 x^2+3 y^2=1\) Then, \(\lambda\) is,

120648 Tangents are drawn to the ellipse \(\frac{x^2}{9}+\frac{y^2}{5}=1\) at the ends of both latus rectum. The area of the quadrilateral, so formed is
#[Qdiff: Hard, QCat: Numerical Based, examname: WB JEE-2022], Where, On comparing standard ellipse equation, \(\frac{\mathrm{x}^2}{\mathrm{~A}^2}+\frac{\mathrm{y}^2}{\mathrm{~B}^2}=1\), \(A^2=36 \& B^2=4\), According to question-, \(\frac{\mathrm{A}^2}{\mathrm{~b}^2}+\frac{\mathrm{B}^2}{\mathrm{a}^2}=\frac{\left(\mathrm{A}^2-\mathrm{B}^2\right)}{\mathrm{c}^2}\), \(\frac{36}{\mathrm{~b}^2}+\frac{4}{\mathrm{a}^2}=\frac{(36-4)^2}{\mathrm{c}^2}\), \(\frac{1}{\mathrm{~b}^2}+\frac{1}{\mathrm{a}^2}=\frac{(32)^2}{2}\), \(\frac{1}{\mathrm{a}^2}+\frac{9}{\mathrm{~b}^2}=\frac{256}{\mathrm{c}^2}\), 921. \(A B\) is a variable chord of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\). If \(A B\) subtends a right angle at the origin \(\mathrm{O}\), then \(\frac{1}{\mathrm{OA}^2}+\frac{1}{\mathrm{OB}^2}\) equals to,

120649 The equation of the tangent to the ellipse \(x^2+\) \(16 \mathrm{y}^2=16\) which makes an angle \(60^{\circ}\) with the \(X\) axis is \(\qquad\)
#[Qdiff: Hard, QCat: Numerical Based, examname: AP EAMCET-22.09.2020,Shift-I],

120651 The eccentricity of an ellipse whose centre is at the origin is \(\mathbf{1 / 2}\). If one of its directrices is \(x=-\) 4 , then the equation of the normal to it at \(\left(1, \frac{3}{2}\right)\) is