The condition for the line I x + m y + n = 0 to be a normal to \; {x²}{25} + \; {y²}{9} = 1 is

Correct Answer is : 9m2+ 25R2= 256n2\; {9}{m²} + \; {25}{R²} = \; {256}{n²}m29+R225=n2256

\; {l²}{9} + \; {m²}{25} = \; {n²}{256}

\; {9}{m²} + \; {25}{R²} = \; {256}{n²}

\; {l²}{9} - \; {m²}{25} = \; {n²}{256}

Correct Answer is : 9m2+ 25R2= 256n2\; {9}{m²} + \; {25}{R²} = \; {256}{n²}m29+R225=n2256

Test Index

Manipal Entrance Test 2020 Math Paper

Question : 16 of 70

Marks: +1, -0

I x + m y + n = 0

\; \frac{x^{2}}{25} + \; \frac{y^{2}}{9} = 1

Solution:

Given equation of line is

I x + m y + n = 0

and equation of ellipse is

\; \frac{x^{2}}{25} + \; \frac{y^{2}}{q} = 1

\therefore

The equation of any normal to the ellipse is

5 x \sec \theta - 3 y \csc \theta = 25 - 9

\begin{pmatrix} \because \; \text{the equation of any normal to the ellipse} \; & \; \frac{x^{2}}{a^{2}} + \; \frac{y^{2}}{b^{2}} = 1 \; \text{ is } \; a x \sec \theta - \; \text{ by } \; \csc \theta = a^{2} - b^{2} \end{pmatrix}

\Rightarrow \; \; 5 x \sec \theta - 3 y \csc \theta - 16 = 0

.....(i)

As the Eq. (i) is the normal to the ellipse.

\therefore \; \; \; \frac{5 \sec \theta}{1} = \; \frac{-3 \csc \theta}{m} = \; \frac{-16}{n}

\Rightarrow \; \; \cos \theta = \; \frac{5 n}{-161}

and

\sin \theta = \; \frac{3 n}{16 m}

\because \; \; \cos^{2} \theta + \sin^{2} \theta = 1

\therefore \; \; \left( \; \frac{5 n}{161} \right)^{2} + \left( \; \frac{3 n}{16 m} \right)^{2} = 1

\Rightarrow \; \; \; \frac{25 n^{2}}{256 I^{2}} + \; \frac{9 n^{2}}{256 m^{2}} = 1

\Rightarrow \; \; \; \frac{25}{l^{2}} + \; \frac{9}{m^{2}} = \; \frac{256}{n^{2}}

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