If the tangent to the curve y = 1 - x^2 at x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $A_1$ be the area under the curve $y = 1 - x^2$ and the axes for $0 < x < 1.$$A_1 = \int_{0}^{1} (1 - x^2) \, dx = \left[ x - \frac{x^3}{3} \r

Vedclass · Accepted Answer

$\frac{2}{\sqrt{3}}$

Vedclass · Answer

(A) Let $A_1$ be the area under the curve $y = 1 - x^2$ and the axes for $0 $A_1 = \int_{0}^{1} (1 - x^2) \, dx = \left[ x - \frac{x^3}{3} ight]_{0}^{1} = 1 - \frac{1}{3} = \frac{2}{3}.$Now,for the curve $y = 1 - x^2,$ the slope of the tangent at $x = \alpha$ is $y' = -2x.$ At $x = \alpha,$ $y' = -2\alpha.$The equation of the tangent at $(\alpha, 1 - \alpha^2)$ is $y - (1 - \alpha^2) = -2\alpha(x - \alpha).$$y - 1 + \alpha^2 = -2\alpha x + 2\alpha^2 \Rightarrow 2\alpha x + y = \alpha^2 + 1.$The tangent meets the $x$-axis at $P$ (where $y=0$) and the $y$-axis at $Q$ (where $x=0$).$P = \left( \frac{\alpha^2 + 1}{2\alpha}, 0 ight)$ and $Q = (0, \alpha^2 + 1).$The area of triangle $OPQ$ is $A = \frac{1}{2} 	imes OP 	imes OQ = \frac{1}{2} 	imes \frac{\alpha^2 + 1}{2\alpha} 	imes (\alpha^2 + 1) = \frac{(\alpha^2 + 1)^2}{4\alpha}.$To find the minimum area,differentiate $A$ with respect to $\alpha$:$A' = \frac{1}{4} \left[ \frac{2(\alpha^2 + 1)(2\alpha)(\alpha) - (\alpha^2 + 1)^2(1)}{\alpha^2} ight] = \frac{(\alpha^2 + 1)(4\alpha^2 - \alpha^2 - 1)}{4\alpha^2} = \frac{(\alpha^2 + 1)(3\alpha^2 - 1)}{4\alpha^2}.$Setting $A' = 0,$ we get $3\alpha^2 - 1 = 0 \Rightarrow \alpha^2 = \frac{1}{3} \Rightarrow \alpha = \frac{1}{\sqrt{3}}$ (since $0 The minimum area $A = \frac{(\frac{1}{3} + 1)^2}{4(\frac{1}{\sqrt{3}})} = \frac{(\frac{4}{3})^2}{\frac{4}{\sqrt{3}}} = \frac{16}{9} 	imes \frac{\sqrt{3}}{4} = \frac{4\sqrt{3}}{9} = \frac{4}{3\sqrt{3}}.$Given $A = k A_1,$ we have $\frac{4}{3\sqrt{3}} = k 	imes \frac{2}{3}.$$k = \frac{4}{3\sqrt{3}} 	imes \frac{3}{2} = \frac{2}{\sqrt{3}}.$

If the tangent to the curve $y = 1 - x^2$ at $x = \alpha,$ where $0 < \alpha < 1,$ meets the axes at $P$ and $Q.$ Also $\alpha$ varies,the minimum value of the area of the triangle $OPQ$ is $k$ times the area bounded by the axes and the part of the curve for which $0 < x < 1,$ then $k$ is equal to

Similar Questions

The area bounded by the parabola $y^{2}=16x$ and its latus rectum in the first quadrant is

Find the area enclosed between the parabola $y^{2}=4ax$ and the line $y=mx$.

Let $S(\alpha) = \{(x,y) : y^2 \leq x, 0 \leq x \leq \alpha\}$ and $A(\alpha)$ be the area of the region $S(\alpha)$. If for a $\lambda, 0 < \lambda < 4, A(\lambda) : A(4) = 2 : 5$,then $\lambda$ equals:

If $x^{2}+y^{2}=a^{2}$, then $\int_{0}^{a} \sqrt{1+\left(\frac{dy}{dx}\right)^{2}} dx=$

The area bounded by the curve $y^2 = x$,the line $y = 4$,and the $y$-axis is

Vedclass Test Series

Exam Paper Generator

Online Exam Module