Solveeit Logo

Question

Mathematics Question on Tangents and Normals

An equation of a common tangent to the parabola y2=163xy^{2} = 16\sqrt{3}x and the ellipse 2x2+y2=42x^{2} + y^{2} = 4 is y=2x+23y = 2x + 2\sqrt{3} If the line y=mx+43m,(m0)y = mx + \frac{4\sqrt{3}}{m}, \left(m\ne0\right) is a common tangent to the parabola y2=163xy^{2} = 16\sqrt{3}x and the ellipse 2x2+y2=42x^{2} + y^{2} = 4, then m satisfies m4+2m2=24.m^{4} + 2m^{2} = 24.

A

Statement-1 is false, Statement-2 is true.

B

Statement-1 is true, statement-2 is true; statement-2 is a correct explanation for Statement-1.

C

Statement-1 is true, statement-2 is true; statement-2 is not a correct explanation for Statement-1.

D

Statement-1 is true, statement-2 is false.

Answer

Statement-1 is true, statement-2 is true; statement-2 is a correct explanation for Statement-1.

Explanation

Solution

Equation of tangent to the ellipse x22+y24=1\frac{x^{2}}{2} + \frac{y^{2}}{4} = 1 is y=mx±2m2+4.....(1)y = mx \pm \sqrt{2m^{2}+4}\quad\quad.....\left(1\right) equation of tangent to the parabola y2=163xy^{2} = 16 \sqrt{3}x is y=mx+43m.....(2)y = m x + \frac{4\sqrt{3}}{m}\quad\quad.....\left(2\right) On comparing (1)\left(1\right) and (2)\left(2\right) 43m=±2m2+4\frac{4\sqrt{3}}{m} = \pm \sqrt{2m^{2}+4} 48=m2(2m2+4)2m4+4m248=0\Rightarrow\quad48 = m^{2} \left(2m^{2} + 4\right)\quad\quad\Rightarrow\quad2m^{4 }+ 4m^{2} - 48 = 0 m4+2m224=0(m2+6)(m24)=0\Rightarrow\quad m^{4 }+ 2m^{2} - 24 = 0\quad\quad\Rightarrow\quad\left(m^{2} + 6\right) \left(m^{2} - 4\right) = 0 m2=4m=±2\Rightarrow\quad m^{2} = 4\quad\quad\Rightarrow\quad m = \pm 2 \Rightarrow\quad equation of common tangents are y=±2x±23y = \pm 2x \pm 2 \sqrt{3} statement -1 is true. statement-2 is obviously true.