If Ax - By -5 is a normal at a point P on the ellipse x 2-9- y 2-4-1 whose eccentric angle is -4- then the value of A - B 2 is

Given: x^29+y^24=1 Here, a=3,b=2 Using parametric form of normal: axcosθ bysinθ=a^2 b^2 ⇒ nbsp;3xcosθ 2ysinθ=5 For θ=π4 the equation of normal is nbsp; ⇒3√ ...

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Standard XII

Mathematics

Parametric Form of Normal : Ellipse

If Ax + By =5...

Question

If $A x + B y = 5$ is a normal at a point $P$ on the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ whose eccentric angle is $\frac{π}{4}$ , then the value of $(A + B)^{2}$ is

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Solution

Given: $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$
Here,
$a = 3, b = 2$
Using parametric form of normal:
$\frac{a x}{cos θ} - \frac{b y}{sin θ} = a^{2} - b^{2} \Rightarrow \frac{3 x}{cos θ} - \frac{2 y}{sin θ} = 5$

For $θ = \frac{π}{4}$ the equation of normal is
$\Rightarrow 3 \sqrt{2} x - 2 \sqrt{2} y = 5$
Comparing with $A x + B y = 5$ , we get
$A = 3 \sqrt{2}, B = - 2 \sqrt{2} ∴ (A + B)^{2} = 2$

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Parametric Form of Normal : Ellipse

Standard XII Mathematics

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If Ax+By=5 is a normal at a point P on the ellipse x29+y24=1 whose eccentric angle is π4, then the value of (A+B)2 is

Given: x29+y24=1 Here, a=3,b=2 Using parametric form of normal: axcosθ−bysinθ=a2−b2⇒3xcosθ−2ysinθ=5 For θ=π4 the equation of normal is ⇒3√2x−2√2y=5 Comparing with Ax+By=5, we get A=3√2, B=−2√2∴(A+B)2=2

If $A x + B y = 5$ is a normal at a point $P$ on the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ whose eccentric angle is $\frac{π}{4}$ , then the value of $(A + B)^{2}$ is