First slide

Ellipse

Question

Statement 1: If the extremities of the latus rectum of the
ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ (a> b), having positive ordinates
lies on the parabola $x^{2} = - 2 (y - 2),$ then, $a = 2$
Statement 2: If the length of the latus rectum of the
ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is equal to the distance between
the foci, then the eccentricity e of the ellipse satisfies
$e^{2} + e - 1 = 0$

Moderate

A

STATEMENT- I is True, STATEMENT-2 is True; S TATEMENT-2 is a correct explanation for STATEMENT-
B

STATEMENT-I is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT- I
C

STATEMENT-I is True, STATEMENT-2 is False
D

STATEMENT-I is False, STATEMENT-2 is True

Solution

Statement-2 is true by definition of conjugate
diameters.
Let $y = m x$ and $y = m^{'} x$ x be two conjugate diameters of
the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 .$ Let $(h, k)$ be the mid point
of chord whose step is m then
$\frac{h x}{a^{2}} + \frac{k y}{b^{2}} - 1 = \frac{h^{2}}{a^{2}} + \frac{k^{2}}{b^{2}} - 1$
$\Rightarrow m = - \frac{b^{2}}{a^{2}} \frac{h}{k} \Rightarrow$ Locus of $(h, k)$ is $y = \frac{- b^{2}}{a^{2} m} x$
$\Rightarrow \frac{- b^{2}}{a^{2} m} = m^{'} \Rightarrow m^{'} m = \frac{- b^{2}}{a^{2}}$ . (using statement-2)
and thus statement- I is false.

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