The ellipse frac{x^2}{a^2}+frac{y^2}{b^2}=1 ( | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the point of intersection be $(x_0, y_0)$. For the parabola $y^2=4ax$,the slope of the tangent $m_1$ at $(x_0, y_0)$ is given by $2y y' = 4a \

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Let the point of intersection be $(x_0, y_0)$. For the parabola $y^2=4ax$,the slope of the tangent $m_1$ at $(x_0, y_0)$ is given by $2y y' = 4a \Rightarrow m_1 = \frac{2a}{y_0}$.For the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,the slope of the tangent $m_2$ at $(x_0, y_0)$ is given by $\frac{2x}{a^2} + \frac{2y y'}{b^2} = 0 \Rightarrow m_2 = -\frac{b^2 x_0}{a^2 y_0}$.Since they intersect at right angles,$m_1 m_2 = -1$,so $\left(\frac{2a}{y_0}\right) \left(-\frac{b^2 x_0}{a^2 y_0}\right) = -1 \Rightarrow \frac{2b^2 x_0}{a y_0^2} = 1$.Substituting $y_0^2 = 4ax_0$,we get $\frac{2b^2 x_0}{a(4ax_0)} = 1$ $\Rightarrow \frac{b^2}{2a^2} = 1$ $\Rightarrow b^2 = 2a^2$.The eccentricity $e$ of the ellipse $(b>a)$ is given by $a^2 = b^2(1-e^2)$ $\Rightarrow a^2 = 2a^2(1-e^2)$ $\Rightarrow 1 = 2(1-e^2)$ $\Rightarrow 1 = 2 - 2e^2$ $\Rightarrow 2e^2 = 1$.

The ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $(b>a)$ and the parabola $y^2=4ax$ intersect at right angles. If $e$ is the eccentricity of the ellipse,then $2e^2=$

Similar Questions

The quadratic equation whose roots are $l$ and $m$,where $l = \lim_{\theta \rightarrow 0} \left( \frac{3 \sin \theta - 4 \sin^2 \theta}{\theta} \right)$ and $m = \lim_{\theta \rightarrow 0} \frac{2 \tan \theta}{\theta(1 - \tan^2 \theta)}$,is:

If $PQ$ is a double ordinate of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ such that $\triangle OPQ$ is an equilateral triangle,where $O$ is the centre of the hyperbola,then the eccentricity $e$ of the hyperbola satisfies:

If the foci of the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$ coincide with the foci of the hyperbola $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$,then $b^2$ is equal to

$A$ common tangent to the conics $x^2 = 6y$ and $2x^2 - 4y^2 = 9$ is

The equations of the common tangents to the two hyperbolas $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module