If m _ 1 ~and& m _ 2 are the slopes of the tangents to the ellipse x^ 2 16 + y^ 2 9 = 1 which passes through (5,4), ~then the value of ( m _ 1 + m _ 2 ) - ( m _ 1 m _ 2 ) is equal to

Correct answer: D. (11)/(3) Let the equation of tangent passing through (5,4) is y = mx ± 16m^ 2 + 9 ⇒ 4 = 5m ± 16m^ 2 + 9 ⇒ (4 - 5m) ^ 2 = 16m^ 2 + 9 ⇒ 9m^ 2 - 40m + 7 = 0 ⇒ m _ 1 + m _ 2 = (40)/(9), m _ 1 m _ 2 = (7)/(9) ⇒ ( m _ 1 + m _ 2 ) - ( m _ 1 m _ 2 ) = (33)/(9) = (11)/(3)

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Ellipse

NTA Abhyas 2020 · Mathematics · Ellipse

If m _ 1 ~and& m _ 2 are the slopes of the tangents to the ellipse...

{m}_{1}

~and&

{m}_{2}

are the slopes of the tangents to the ellipse

\dfrac{x^{2}}{16} + \dfrac{y^{2}}{9} = 1

which passes through \left(5,4\right),

~then the value of \left({m}_{1} + {m}_{2}\right) - \left({m}_{1}{m}_{2}\right) is equal to

Options

A $\dfrac{47}{9}$
B $- \dfrac{40}{9}$
C $\dfrac{22}{3}$
D $\dfrac{11}{3}$

Correct answer

D. $\dfrac{11}{3}$

Step-by-step solution

Let the equation of tangent passing through

\left(5,4\right)

y = mx \pm \sqrt{16m^{2} + 9}

⇒ 4 = 5m \pm \sqrt{16m^{2} + 9}

⇒ {\left(4 - 5m\right)}^{2} = 16m^{2} + 9 ⇒ 9m^{2} - 40m + 7 = 0

⇒ {m}_{1} + {m}_{2} = \dfrac{40}{9}, {m}_{1}{m}_{2} = \dfrac{7}{9}

⇒ \left({m}_{1} + {m}_{2}\right) - \left({m}_{1}{m}_{2}\right) = \dfrac{33}{9} = \dfrac{11}{3}

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