Question

Find the condition for the line lx+my+n=0 to be a normal to the ellipse $\mathrm{\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1}$​

Answer 1

$\bigstar\bf\underline{\underline{Given:-}}$

• $\displaystyle{lx + my + n = 0}$
• $\displaystyle{\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1}$

$\bigstar\bf\underline{\underline{To \ find:-}}$

• The condition of line $\displaystyle{lx + my + n = 0}$

$\bigstar\bf\underline{\underline{Answer:-}}$

Let,

$\displaystyle{\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1}$

$\displaystyle{lx + my + n = 0}$

$\displaystyle{y = -\frac{l}{m}\times -\frac{n}{m}}$

Equation of normal

$\implies\displaystyle{y = mx\mp \frac{(a^2-b^2)m}{\sqrt{a^2 + b^2m^2}}}$

$\therefore\displaystyle{-\frac{1}{m} = m\mp \frac{(a^2-b^2)m}{\sqrt{a^2 + b^2(m^2)}} = -\frac{n}{m}}$

$\therefore\displaystyle{\frac{n}{m} =\mp \frac{(a^2 - b^2)\frac{l}{m}}{\sqrt{a^2 + b^2\times \frac{l^2}{m}}}}$

$\implies\displaystyle{\frac{\sqrt{a^2 m^2 + b^2 l^2}}{m} = \mp (a^2 + b^2)\frac{l}{n}}$

$\therefore\displaystyle{a^2 m^2 + b^2 l^2 = (a^2 - b^2)^2\frac{l^2 m^2}{n^2}}$

$\implies\displaystyle{\frac{a^2}{l^2} + \frac{b^2}{m^2} = \boxed{\bf\frac{(a^2 - b^2)^2}{n^2}}}$

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Answer 2

Question:Find the condition for the line lx+my+n=0 is normal to the hyperbola a2x2−b2y2=1 .

As we know that

Equation of normal with slope p to a2x2−b2y2=1 is y=px±a2−b2p2p(a2+b2)

Given lx+my+n=0 is the normal

⟹y=−mlx−mn

⟹p=−ml,

And also

±a2−p2b2p(a2+b2)=−mn

Squaring on both sides

⟹p2(a2+b2)=m2n2(a2−pb2)

⟹m2l2(a2+b2)=m2n2(a2−m2l2b2)

⟹l2m2(a2+b2)=n2(a2m2−l2b2)

Hope it helps ❤️❤️❤️