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Find the equation of the tangent and normal to the curve $16 x^{2} + y^{2} = 145$ at point $(x_{1}, y_{1})$ , where $x_{1} = 2$ and $y_{1} > 0$ .

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3 y - 5 x = 32

45 y - 6 x = 30

5 y - 7 x = 130

32 y - 9 x = 270

answer is D.

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Detailed Solution

16 x^{2} + y^{2} = 145

\Rightarrow 16 \times 2^{2} + y^{2} = 145

\Rightarrow 64 + y^{2} = 145

\Rightarrow y^{2} = 81

\Rightarrow y = 9, \neq - 9

Since, it is given that

y_{1} > 0, y = 9

∴ (x_{1}, y_{1}) = (2, 9)

Finding the slope of tangent at

(x_{1}, y_{1}) = (2, 9)

Let us find the derivative of

16 x^{2} + y^{2} = 145

(x_{1}, y_{1}) = (2, 9)

.
Differentiating the above function with respect to

x

, we get,

\Rightarrow \frac{dy}{dx} (16 x^{2} + y^{2} = 145)

\Rightarrow \frac{dy}{dx} (16 x^{2} + y^{2}) = \frac{dy}{dx} (145)

\Rightarrow 32 x + 2 y \frac{dy}{dx} = 0

\Rightarrow \frac{dy}{dx} = \frac{- 32 x}{2 y}

Now, we will substitute

(x_{1}, y_{1}) = (2, 9)

in the equation.

\Rightarrow \frac{dy}{dx} = \frac{- 32 x}{2 y}

\Rightarrow \frac{dy}{dx} = \frac{- 32 \times 2}{2 \times 9}

\Rightarrow \frac{dy}{dx} = \frac{- 32}{9}

The slope of tangent can be written as

m_{t} = \frac{- 32}{9}

.
Finding the equation of tangent,

(x_{1}, y_{1})

is given as

y - y_{1} = m_{t} (x - x_{1})

Substituting

y - 9 = \frac{- 32}{9} (x - 2)

\Rightarrow y - 9 = \frac{- 32 x + 64}{9}

\Rightarrow 9 y - 81 = - 32 x + 64

\Rightarrow 9 y + 32 x = 145

Hence, the equation of tangent at

(x_{1}, y_{1}) = (2, 9)

9 y + 32 x = 145

.
Finding the slope of normal
Substituting the value of

m_{t} = \frac{- 32}{9}

,
We get

m_{n} \times m_{t} = - 1

\Rightarrow m_{n} \times \frac{- 32}{9} = - 1

\Rightarrow m_{n} = \frac{9}{32}

Equation of a normal with slope

m_{n}

at point

(x_{1}, y_{1})

is given as

y - y_{1} = m_{n} (x - x_{1})

\Rightarrow y - 9 = \frac{9}{32} (x - 2)

\Rightarrow y - 9 = \frac{9 x - 18}{32}

\Rightarrow 32 y - 288 = 9 x - 18

\Rightarrow 32 y - 9 x = 270

Hence, the equation of normal at

(x_{1}, y_{1}) = (2, 9)

32 y - 9 x = 270

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