JEE Main 2022 · 25 Jun · Shift 2 · Mathematics Q29

← JEE Main 2022 · 25 Jun · Shift 2

MathematicsInteger+4 / -1Question 29 of 90

Let the eccentricity of the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ be $\dfrac54$ . If the equation of the normal at the point $\left(\dfrac8{\sqrt5},\dfrac{12}5\right)$ on the hyperbola is $8\sqrt5x+\beta y=\lambda$ , then $\lambda-\beta$ is equal to ______.

Your answer

Answer and worked solution

Official answer: 85

Shape. $e=\frac54$ gives $b^2=\frac9{16}a^2$ ; the point gives $\dfrac{64}{5a^2}-\dfrac{144}{25b^2}=1$ , so $a^2=\frac{64}{25}$ , $b^2=\frac{36}{25}$ .

Normal. $\dfrac{a^2x}{x_1}+\dfrac{b^2y}{y_1}=a^2+b^2$ : $\dfrac{8\sqrt5}{25}x+\dfrac35y=4$ , i.e. $8\sqrt5x+15y=100$ .

Value. $\lambda-\beta=100-15=85$ .

Answer: 85

Topics:Hyperbola Tangents and normals to conics

← Previous question Next question →

JEE Main 2022 · 25 Jun · Shift 2, Mathematics question 29