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NIMCET 2016 — Mathematics PYQ

NIMCET | Mathematics | 2016

The line $3 x + 5 y = k$ touches the ellipse $16 x^{2} + 25 y^{2} = 400$ if k is

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Explanation

Concept:
The standard form of the equation of ellipse:
$\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$
where center coordinates are (0, 0),
a = length of semi-major axis, and
b = length of semi-minor axis
For a line y = mx + c to be tangent of such ellipse
where m is the slope of the line and c is a constant, then: c^2 = a^2m^2 + b^2 condition must follow, i.e.,
y = mx ± √(a^2m^2 + b^2) is the standard form of tangent to $< / s p an >< b r >< s p an s t y l e = " f o n t - s i z e : 14 pt; " > \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ ellipse

Calculation:
The given ellipse equation:
$16 x^{2} + 25 y^{2} = 400$
It can be rewritten in standard form as,
$\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1$
$\Rightarrow a = 25 = 5$
$\Rightarrow b = 16 = 4$
The equation of a line given:
$3 x + 5 y = k,$
$\Rightarrow y = \frac{- 3}{5} x + \frac{k}{5}$
$∴ m = \frac{- 3}{5}, c = \frac{k}{5}$
For the line to be tangent,
$c^{2} = a^{2} m^{2} + b^{2}$
$\Rightarrow (\frac{k}{5})^{2} = 5^{2} (\frac{- 3}{5})^{2} + 4^{2}$
$\Rightarrow \frac{k ^{2}}{25} = 9 + 16 = 25$
$\Rightarrow k^{2} = 625$
$\Rightarrow k = \pm 25$

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