Question

show that the line x-y = 5 is a tangent to the ellipse 9x^+16y^ = 144 find the point of contact.​

Answer 1

$Given \: Equation \: 9x^{2} + 16y^{2} = 144$

/* Dividing each term by 144, we get */

$\implies \frac{9x^{2}}{144} + \frac{16y^{2}}{144} = \frac{144}{144}$

$\implies \frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$

$a^{2} = 16,\: b^{2} = 9$

$Given \: tangent \: x - y = 5 .$

$\implies y = x - 5 \: Compare \:this \:with \\y = mx + c , we \:get$

$m = 1 \: and \: c = - 5$

/* The condition that the line y = mx+c may be a tangent to the ellipse

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}= 1 \\is \: c^{2} = a^{2}m^{2} + b^{2}$

$\implies (-5)^{2} = 16 \times 1^{2} + 9$

$\implies 25 = 16 + 9$

$Here, c = -5 \: but \: c \neq 5$

$Now, point \:of \: contact ( x_{1} , y_{1} ) \\= \Big( \frac{-a^{2}m}{c}, \frac{b^{2}}{c}\Big)$

$= \Big( \frac{-16 \times 1 }{-5} , \frac{9}{-5}\Big) \\= \Big( \frac{16}{5} , \frac{-9}{5}\Big)$

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