First slide

Hyperbola in conic sections

Question

Statement-1: Two tangents drawn from any point on he hyperbola $x^{2} - y^{2} = a^{2} - b^{2} to the ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$ =1 make complementary angles with the axis of the ellipse.
Statement-2: If two lines make complementary angles with the axis of x then the product of their slopes is 1.

Moderate

A

STATEMENT-1 is True, STATEMENT-2 is True;
STATEMENT-2 is a correct explanation for STATEMENT-1
B

STATEMENT-1 is True, STATEMENT-2 is True;
STATEMENT-2 is NOT a correct explanation for STATEMENT-1
C

STATEMENT-1 is True, STATEMENT-2 is False
D

STATEMENT-1 is False, STATEMENT-2 is True

Solution

If the angle in statement-2 are $α and β$ such that $α + β = π / 2$ then the product of the slopes is tan $α \tan β$ = 1, and the statement-2 is true.
In statement-1, any tangent to the ellipse is
$y = m x + \sqrt{a^{2} m^{2} + b^{2}}$ which passes through
$\begin{array}{l} (\sqrt{a^{2} - b^{2}} \sec θ, \sqrt{a^{2} - b^{2}} \tan θ) \\ \Rightarrow (a^{2} - b^{2}) (\tan θ - m \sec θ)^{2} = a^{2} m^{2} + b^{2} \end{array}$
Product of the slopes
$= \frac{(a^{2} - b^{2}) \tan^{2} θ - b^{2}}{(a^{2} - b^{2}) \sec^{2} θ - a^{2}} = \frac{a^{2} \sin^{2} θ - b^{2}}{a^{2} \sin^{2} θ - b^{2}} = 1$
So by statement-2, statement-1 is also true.

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