Question

please help me


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Answer 1

Answer:

Option (1)

Note:

• The first derivative of the equation of any curve give the slope of the tangent.

ie; slope of tangent = dy/dx.

• The first derivative of the equation at a particular point will give the slope of the tangent at that particular point.

• The product of slopes of two mutually perpendicular lines is -1.

• For any curve,

(slope of tangent)(slope of normal)= -1

OR

slope of normal = -1/slope of tangent

OR

slope of normal = -1/(dy/dx) = -(dx/dy)

• The Point-slope form of a straight line is given by ; m = (y-y1)/(x-x1)

where, "m" is the slope of the straight line and (x1,y1) is the given point which lies on that line.

{ for explaination, please refer to the attachments }

Answer 2

Answer:-

$\implies \: \boxed{\frac{y {b}^{2} }{ y_1 } - \frac{x {a}^{2} }{ x_1} = {b}^{2} - {a}^{2} }$

Option (1) is correct.

Step - by - step explanation:-

To find :-

From nd equation of normal to the given curve at given points.

Solution :-

Given curve →

$\frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} } = 1 \\ \\ \bf{differentiating \: with \: respect \: to \:} x \\ \\ \frac{2x}{ {a}^{2} } + \frac{2y}{ {b}^{2} } \frac{dy}{dx} = 0 \\ \\ \frac{dy}{dx} = - \frac{x \: {b}^{2} }{y \: {a}^{2} } \\ \\ \therefore \: slope \: of \: normal \: at \: ( \: x_1\: y_1 \:) \: </em><em>is\</em><em>:</em></p><p><em>m_n\: \: \\ \\ \implies \: \frac{dy}{dx} | at \: ( \: x_1\: y_1 \:) \: = \frac{ - 1}{ \frac{dy}{dx} } \\ \\ \implies \: m_n = \frac{y_1 \: {a}^{2} }{ x_1 \: {b}^{2} } \\ \\ hence \\ \\ equation \: of \: normal \: \downarrow \\ \\ \implies \: y - y_1 = m_n(x - x_1) \\ \\ \implies \: y - y_1\: = \frac{ \: y_1 {a}^{2} \: }{ x_1 \: {b}^{2} } \bigg(x - x_1 \bigg) \\ \\ \implies \: y \: x_1 \: {b}^{2} - x_1y_1 {b}^{2} = x y_1 \: {a}^{2} - x_1\: y_1 {a}^{2} \\ \\ \implies \: y \: x_1 \: {b}^{2} \: - x y_1 \: {a}^{2} \: = x_1y_1 {b}^{2} - x_1\: y_1 {a}^{2} \\ \\ \implies \: y \: x_1 \: {b}^{2} \: - x y_1 \: {a}^{2} \: \: = x_1 y_1 ( {b}^{2} \: - {a}^{2} ) \\ \\ \implies \: \frac{y {b}^{2} }{ y_1 } - \frac{x {a}^{2} }{ x_1} = {b}^{2} - {a}^{2}$

Hope it helps you.