The tangents at the point $A (1,3)$ and $B (1,-1)$ on the parabola $y^2-2 x-2 y=1$ meet at the point $P$ Then the area (in unit $^2$ ) of $\triangle PAB$ is :-
- 4
- 6
- 7
- 8
The Correct Option is D
Solution and Explanation
Equation of tangent aty is
and equation of tangent at is
⇒So point is
⇒ A = $\frac1{2}$ $ \begin{bmatrix} 1 & 3 & 1 \\ 1 & -1 & 1 \\ -3 & 1 & 1 \end{bmatrix} =8$
Therefore, the correct answer is option (D) : 8
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Concepts Used:
Application of Derivatives
Various Applications of Derivatives-
Rate of Change of Quantities:
If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by
\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)
This is also known to be as the Average Rate of Change.
Increasing and Decreasing Function:
Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).
- If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
- Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
- The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.
Read More: Application of Derivatives
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