6. ⇒ (MHT CET 2023 9th May Evening Shift)

If slope of a tangent to the curve $x y + a x + b y = 0$ at the point $(1, 1)$ on it is 2 ,

A. 3

B. 1

C. 2

D. $-$ 1

Correct Option is (A)

Given curve is

$\begin{array}{ll} x y + a x + b y = 0 \\ ∴ & Slope = 2 = \frac{d y}{d x} \\ ∴ & x y + a x + b y = 0 \end{array}$

Differentiating w.r.t. $x$ , we get

$\begin{array}{ll} x \frac{d y}{d x} + y + a + b \frac{d y}{d x} = 0 \\ ∴ & (x + b) \frac{d y}{d x} = - (y + a) \\ \frac{d y}{d x} = \frac{- (y + a)}{x + b} \\ ∴ & {(\frac{d y}{d x})}_{(1, 1)} = 2 \\ ∴ & 2 = \frac{- (1 + a)}{1 + b} \\ ∴ & a + 2 b = - 3 ..... (i) \end{array}$

$\begin{aligned} Since (1, 1) lies on x y + a x + b y = 0, we get \\ a + b = - 1 .... (ii) \end{aligned}$

Solving (i), (ii), we get

$\begin{aligned} a = 1, b = - 2 \\ ∴ & a - b = 1 - (- 2) = 3 \end{aligned}$

7. ⇒ (MHT CET 2023 9th May Morning Shift )

A spherical raindrop evaporates at a rate proportional to its surface area. If originally its radius is $3 mm$ and 1 hour later it reduces to $2 mm$ , then the expression for the radius $R$ of the raindrop at any time $t$ is

A. $6 R = t + 2$

B. $R (t + 2) = 6$

C. $R = 6 (t + 2)$

D. $6 R = t$

Correct Option is (B)

According to the given conditions, when $t = 0, R = 3$ and when $t = 1, R = 2$

This condition is satisfied by only option (B)

8. ⇒ (MHT CET 2021 21th September Morning Shift )

The curves $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{4} = 1$ and $y^{3} = 16 x$ intersect each other orthogonally, then $a^{2} =$

A. 2

B. $\frac{3}{4}$

C. $\frac{1}{2}$

D. $\frac{4}{3}$

Correct Option is (D)

$\begin{aligned} \frac{x^{2}}{a^{2}} + \frac{y^{2}}{4} = 1 \\ ∴ \frac{1}{a^{2}} 2 x + \frac{1}{4} 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = (\frac{- x}{a^{2}}) (\frac{4}{y}) .....(1) \end{aligned}$

Also $y^{3} = 16 x$

$∴ 3 y^{2} \frac{d y}{d x} = 16 \Rightarrow \frac{d y}{d x} = \frac{16}{3 y^{2}}$ ...... (2)

Since curves intersect orthogonally, from (1) and (2), we write

$\begin{aligned} (\frac{- x}{a^{2}}) (\frac{4}{y}) (\frac{16}{3 y^{2}}) = - 1 \\ ∴ \frac{64 x}{3 a^{2} y^{3}} = 1 and we have y^{3} = 16 x \\ ∴ \frac{64 x}{3 a^{2} (16 x)} = 1 \Rightarrow a^{2} = \frac{4}{3} \end{aligned}$

9. ⇒ (MHT CET 2021 20th September Evening Shift )

The equation of tangent to the circle $x^{2} + y^{2} = 64$ at the point $P (\frac{2 π}{3})$ is

A. $x - \sqrt{3} y - 16 = 0$

B. $\sqrt{3} x + y - 16 = 0$

C. $x + \sqrt{3} y + 16 = 0$

D. $x - \sqrt{3} y + 16 = 0$

Correct Option is (D)

Circle $x^{2} + y^{2} = (8)^{2}$ , has radius 8 and centre $(0, 0)$ . Point $P (\frac{2 π}{3})$ on the circle has coordinates

$P \equiv (8 \cos \frac{2 π}{3}, 8 \sin \frac{2 π}{3}) i.e. P \equiv (- 4, 4 \sqrt{3})$

Differentiating equation of circle w.r.t. $x$ , we get

$2 x + 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{- x}{y} \Rightarrow {(\frac{d y}{d x})}_{P} = \frac{4}{4 \sqrt{3}} = \frac{1}{\sqrt{3}}$

Hence required equation of tangent is

$(y - 4 \sqrt{3}) = \frac{1}{\sqrt{3}} (x + 4) \Rightarrow x - \sqrt{3} y + 16 = 0$

10. ⇒ (MHT CET 2021 20th September Morning Shift )

The equation of the tangent to the curve $y = 4 x e^{x}$ at $(- 1, \frac{- 4}{e})$ is

A. $6 x - \frac{e}{4} y = - 5$

B. $x - \frac{e}{4} y = 0$

C. $x = - 1$

D. $y = \frac{- 4}{e}$

Correct Option is (D)

$\begin{aligned} y = 4 x e^{x} \\ ∴ \frac{d y}{d x} = 4 x e^{x} + 4 e^{x} \\ ∴ {(\frac{d y}{d x})}_{(- 1, \frac{- 4}{e})} = 4 (- 1) e^{- 1} + 4 e^{- 1} = \frac{- 4}{e} + \frac{4}{e} = 0 \end{aligned}$

Thus tangent is parallel to $X$ axis.

Hence required equation of tangent is $y = \frac{- 4}{e}$

⇒Application Of Derivative