Question

f the line $y=4x-5$ touches to the curve ${{y}^{2}}=a{{x}^{3}}+b$ at the point (2,3) then $7a+2b$
A.0
B.1
C.-1
D.2

Answer 1

Hint: In this question, the line $y=4x-5$ touches the curve ${{y}^{2}}=a{{x}^{3}}+b$ at the point (2,3). It means that the line $y=4x-5$ is tangent to the curve ${{y}^{2}}=a{{x}^{3}}+b$ at the point (2,3). The slope of the line $y=4x-5$ is 4. Also, the slope of the tangent to the curve is given by differentiating the curve ${{y}^{2}}=a{{x}^{3}}+b$ with respect to x. So, the slope of the tangent will be $m=\dfrac{3a{{x}^{2}}}{2y}$ . The point (2,3) is on the curve ${{y}^{2}}=a{{x}^{3}}+b$ , so the coordinates of the point (2,3) must satisfy the equation of the curve ${{y}^{2}}=a{{x}^{3}}+b$ . Now, we have two equations and two variables and solve it further.

Complete step-by-step answer:

According to the question, it is given that the line $y=4x-5$ touches the curve ${{y}^{2}}=a{{x}^{3}}+b$ at the point (2,3).
It means that the line $y=4x-5$ is tangent to the curve ${{y}^{2}}=a{{x}^{3}}+b$ at the point (2,3).
$y=4x-5$ ………………..(1)
${{y}^{2}}=a{{x}^{3}}+b$ ……………………….(2)
The slope of the line $y=4x-5$ which is also a tangent to the curve ${{y}^{2}}=a{{x}^{3}}+b$ is 4.
$m=4$ ……………(3)
The slope of the tangent to the curve is given by differentiating the curve with respect to x.
Now differentiating the curve ${{y}^{2}}=a{{x}^{3}}+b$ with respect to x, we get
We know the formula, $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}$ . Now, using this formula differentiate the curve ${{y}^{2}}=a{{x}^{3}}+b$ , we get
${{y}^{2}}=a{{x}^{3}}+b$
$\Rightarrow 2y\dfrac{dy}{dx}=3a{{x}^{2}}$
$\Rightarrow 2y.m=3a{{x}^{2}}$ , where m is the slope of the tangent.
As the line is tangent at point(2,3), so tangent at this point will satisfy the coordinates of this point.
$\Rightarrow 2y.m=3a{{x}^{2}}$

& \Rightarrow 2.3.m=3a{{2}^{2}} \\\ & \Rightarrow 6m=12a \\\ \end{aligned}$$ $$\Rightarrow m=2a$$ ………………..(4) In equation (3), we have the value of the slope of the tangent. From equation (3) and equation (4), we have $$2a=4$$ $$\Rightarrow a=2$$ ………………………….(5) As the point (2,3) is on the curve, so the coordinates of this point must satisfy the equation of the curve. Putting x=2 and y=3 in the equation (2), we get $${{y}^{2}}=a{{x}^{3}}+b$$ $$\begin{aligned} & \Rightarrow {{3}^{2}}=a{{2}^{3}}+b \\\ & \Rightarrow 9=8a+b \\\ \end{aligned}$$ Now, putting the value of from equation (5) in the above equation we get $$\begin{aligned} & \Rightarrow 9=8a+b \\\ & \Rightarrow 9=8.2+b \\\ & \Rightarrow 9=16+b \\\ \end{aligned}$$ $$\Rightarrow -7=b$$ …………………………(6) We have to find the value of $$7a+2b$$ . For that, we have to put the values of a and b in the equation $$7a+2b$$ . Now, putting the values of a and b from equation (5) and equation (6) in the equation $$7a+2b$$ , we get $$\begin{aligned} & 7a+2b \\\ & =7.2+2.(-7) \\\ & =14-14 \\\ & =0 \\\ \end{aligned}$$ So, the value of $$7a+2b$$ is 0. Hence, the correct option is A. Note: In this question, one can miss a piece of hidden information. Here, the hidden information is that the line $$y=4x-5$$ is tangent to the curve $${{y}^{2}}=a{{x}^{3}}+b$$ . In the question, it is given that the line touches the curve $${{y}^{2}}=a{{x}^{3}}+b$$ at one point, and also a tangent touches the curve at one point. Therefore, the line $$y=4x-5$$ is tangent to the curve $${{y}^{2}}=a{{x}^{3}}+b$$ .