The angle between the tangents drawn from a point (-3,2) to the ellipse 4 x²+9 y²-36=0 is [AP EAPCET 22 May 2025 S1]

Correct Answer is : 90∘90^{}90∘

Correct Answer is : 90∘90^{}90∘

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Section: Mathematics

Question : 5 of 16

Marks: +1, -0

(-3,2)

4 x^2+9 y^2-36=0

Solution:

Given equation of ellipse

4 x^2+9 y^2-36\;=0

\Rightarrow \;\; \;\frac{x^2}{9}+\;\frac{y^2}{4}\;=1

Equation of pair of tangent from

P (-3,2)

\; \;\; S S_1=T^2

\; \left(\;\frac{x^2}{9}+\;\frac{y^2}{4}-1\right) \;\left(\frac{(-3)^2}{9}+\frac{(2)^2}{4}-1\right)

\; =\left(-\frac{3(x)}{9}+\frac{2(y)}{4}-1\right)^2

\;\Rightarrow \;\frac{x^2}{9}+\;\frac{y^2}{4}-1=\left(-\frac{x}{3}+\frac{y}{2}-1\right)^2

\;\Rightarrow \;\frac{x^2}{9}+\;\frac{y^2}{4}-1=\frac{x^2}{9}+\frac{y^2}{4}-\frac{xy}{3}-y+\frac{2x}{3}

\;\Rightarrow \;\frac{xy}{3}-\frac{2x}{3}+y-2=0

\;\Rightarrow x y-2 x+3 y-6=0

\;\Rightarrow (x+3)(y-2)=0

\;\Rightarrow x+3=0 \;\text{ and }\; y-2=0

\;\Rightarrow x=-3 \;\text{ and }\; y=2

The tangent lines are

x=-3

(a vertical line) and

y=2

(a horizontal line)

\therefore

These ar perpendicular

\therefore \;\; \theta = 90^{\circ}

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