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.
\[\Rightarrow 2y.m=3a{{x}^{2}}\]
\[\begin{align}
  & \Rightarrow 2.3.m=3a{{2}^{2}} \\
 & \Rightarrow 6m=12a \\
\end{align}\]
\[\Rightarrow m=2a\] ………………..(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
\[{{y}^{2}}=a{{x}^{3}}+b\]
\[\begin{align}
  & \Rightarrow {{3}^{2}}=a{{2}^{3}}+b \\
 & \Rightarrow 9=8a+b \\
\end{align}\]
Now, putting the value of from equation (5) in the above equation we get
\[\begin{align}
  & \Rightarrow 9=8a+b \\
 & \Rightarrow 9=8.2+b \\
 & \Rightarrow 9=16+b \\
\end{align}\]
\[\Rightarrow -7=b\] …………………………(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
\[\begin{align}
  & 7a+2b \\
 & =7.2+2.(-7) \\
 & =14-14 \\
 & =0 \\
\end{align}\]
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 \[{{y}^{2}}=a{{x}^{3}}+b\] .