Question

14. Find the equation of tangent to the parabola
(ii) y2 = 36x from the point (2,9)​

Answer 1

EXPLANATION.

⇒ y² = 36x.

From point = (2,9).

As we know that,

Equation of parabola,

⇒ y² = 4ax.

Compare the equation with general equation.

⇒ 4a = 36.

⇒ a = 36/4.

⇒ a = 9.

Slope form : The equation of the tangent of the parabola y² = 4ax is,

y = mx + a/m.

Put the value in equation, we get.

⇒ 9 = m(2) + 9/m.

⇒ 9 = 2m + 9/m.

⇒ 9 = 2m² + 9/m.

⇒ 9m = 2m² + 9.

⇒ 2m² - 9m + 9 = 0.

⇒ 2m² - 6m - 3m + 9 = 0.

⇒ 2m(m - 3) - 3(m - 3) = 0.

⇒ (2m - 3)(m - 3) = 0.

⇒ m = 3/2  and  m = 3.

As we know that,

Equation of tangent.

⇒ (y - y₁) = m(x - x₁).

Slope = 3 = M₁.

Passes through the point = (2,9).

⇒ (y - 9) = 3(x - 2).

⇒ y - 9 = 3x - 6.

⇒ y - 3x - 3 = 0.

Slope = 3/2 = M₂.

Passes through the point = (2,9).

⇒ (y - 9) = 3/2(x - 2).

⇒ 2(y - 9) = 3(x - 2).

⇒ 2y - 18 = 3x - 6.

⇒ 2y - 3x - 12 = 0.

                                                                                                                         

MORE INFORMATION.

Conditions of normal.

The line y = mx + c is a normal to the parabola.

(1) = y² = 4ax if c = -2am - am³.

(2) = x² = 4ay if c = 2a + a/m².

Answer 2

• Let the point (2, 9) represents the coordinate K.

Let the point where the tangent touch the curve y² = 36x be P(h, k).

• Since the point (h, k) lies on y² = 36x.

• ⇛ k² = 36h -----(1)

Now slope of tangent joining the points K(2, 9) and P(h, k) is given by

$\rm :\implies\: \boxed{ \pink{ \bf \: Slope \: of \: PK \: = \tt \:\dfrac{k - 9}{h - 2} }} - - - (2)$

Now,

Consider the curve

$\rm :\implies\: {y}^{2} = 36$

On differentiating both sides w. r. t. x, we get

$\rm :\implies\:2y\dfrac{dy}{dx} = 36$

$\rm :\implies\:\dfrac{dy}{dx} = \dfrac{18}{y}$

$\rm :\implies\: \boxed{ \pink{ \bf \:Slope \: of \: PK \: at \: (h, k)\: = \tt \: \dfrac{18}{k} }} - - - (3)$

Now,

On equating equation (2) and (3), we get

$\rm :\implies\:\dfrac{k - 9}{h - 2} = \dfrac{18}{k}$

$\rm :\implies\: {k}^{2} - 9k = 18h - 36$

$\rm :\implies\: {k}^{2} - 9k = 18 \times \dfrac{ {k}^{2} }{36} - 36$

$\rm :\implies\: {k}^{2} - 9k = \dfrac{ {k}^{2} }{2} - 36$

$\rm :\implies\: {2k}^{2} - 18k = {k}^{2} - 72$

$\rm :\implies\: {k}^{2} - 18k + 72 = 0$

$\rm :\implies\: {k}^{2} - 12k - 6k + 72 = 0$

$\rm :\implies\:k(k - 12) - 6(k - 12) = 0$

$\rm :\implies\:(k - 12)(k - 6) = 0$

$\rm :\implies\:k = 12 \: \: or \: \: k = 6$

Now,

• As k² = 36h,

So,

• When k = 12, ⇛ h = 4

and

• When k = 6 ⇛ h = 1.

• So, point of contact of tangent drawn from the point (2, 9) are (4, 12) and (1, 6).

So, equation of tangent,

Case - 1.

Equation of tangent pssses through two points (2, 9) and (4, 12) is given by

$\rm :\implies\:y - 9 = \dfrac{12 - 9}{4 - 2} (x - 2)$

$\tt \because \: \bf \{\:y - y_1 = \dfrac{y_2-y_1}{x_2-x_1} (x-x_1) \}$

$\rm :\implies\:2y - 18 = 3x - 6$

$\boxed{ \red{\rm :\implies\:3x - 2y + 12 = 0}}$

Case - 2.

Equation of tangent pssses through two points (2, 9) and (1, 6) is given by

$\rm :\implies\:y - 9 = \dfrac{6 - 9}{1 - 2} (x - 2)$

$\rm :\implies\:y - 9 = 3x - 6$

$\boxed{ \red{ \rm :\implies\:3x - y + 3 = 0}}$