Let a, b and lambda be positive real numbers. | vedclass

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(A) The parabola is $y^2 = 4 \lambda x$. The end point of the latus rectum is $P(\lambda, 2 \lambda)$.The slope of the tangent to the parabola at $P$

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}}$

Vedclass · Answer

(A) The parabola is $y^2 = 4 \lambda x$. The end point of the latus rectum is $P(\lambda, 2 \lambda)$.The slope of the tangent to the parabola at $P$ is given by differentiating $y^2 = 4 \lambda x$ with respect to $x$: $2y \frac{dy}{dx} = 4 \lambda \Rightarrow \frac{dy}{dx} = \frac{2 \lambda}{y}$.At $P(\lambda, 2 \lambda)$,the slope $m_1 = \frac{2 \lambda}{2 \lambda} = 1$.The slope of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at $P$ is found by differentiating: $\frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{b^2 x}{a^2 y}$.At $P(\lambda, 2 \lambda)$,the slope $m_2 = -\frac{b^2 \lambda}{a^2 (2 \lambda)} = -\frac{b^2}{2a^2}$.Since the tangents are perpendicular,$m_1 	imes m_2 = -1$.$1 	imes (-\frac{b^2}{2a^2}) = -1$ $\Rightarrow \frac{b^2}{2a^2} = 1$ $\Rightarrow \frac{b^2}{a^2} = 2$.Since the ellipse passes through $P(\lambda, 2 \lambda)$,we have $\frac{\lambda^2}{a^2} + \frac{4 \lambda^2}{b^2} = 1$.Substituting $b^2 = 2a^2$: $\frac{\lambda^2}{a^2} + \frac{4 \lambda^2}{2a^2} = 1$ $\Rightarrow \frac{\lambda^2}{a^2} + \frac{2 \lambda^2}{a^2} = 1$ $\Rightarrow \frac{3 \lambda^2}{a^2} = 1$ $\Rightarrow a^2 = 3 \lambda^2$.Then $b^2 = 2a^2 = 6 \lambda^2$.The eccentricity $e$ is given by $e^2 = 1 - \frac{b^2}{a^2}$ (if $a  b$).Here $\frac{b^2}{a^2} = 2$,so $a^2 Therefore,$e = \frac{1}{\sqrt{2}}$.

List-$I$	List-$II$
$(A)$ $\left[\frac{p}{2}\left(t+\frac{1}{t}\right), \frac{q}{2}\left(t-\frac{1}{t}\right)\right]$	$(I)$ parabola
$(B)$ $(p+q \cos \theta, r+q \sin \theta)$	$(II)$ circle
$(C)$ $(p+\lambda^2, q-\lambda)$	$(III)$ ellipse
	$(IV)$ hyperbola

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