Find the area of the triangle formed by the X | vedclass

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

Question

(C) Given the ellipse equation: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \dots (i)$To find the equation of the tangent and normal at the point $A\left(\frac

Vedclass · Accepted Answer

$\frac{b}{4 a}\left(a^2+b^2\right)$

Vedclass · Answer

(C) Given the ellipse equation: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \dots (i)$To find the equation of the tangent and normal at the point $A\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}ight)$,we first find the derivative $\frac{dy}{dx}$.Differentiating $(i)$ with respect to $x$:$\frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{b^2 x}{a^2 y}$.At point $A$,the slope of the tangent $m_T = -\frac{b^2}{a^2} \left( \frac{a/\sqrt{2}}{b/\sqrt{2}} ight) = -\frac{b}{a}$.The slope of the normal $m_N = -\frac{1}{m_T} = \frac{a}{b}$.Equation of tangent at $A$: $y - \frac{b}{\sqrt{2}} = -\frac{b}{a} \left( x - \frac{a}{\sqrt{2}} ight) \dots (ii)$.Equation of normal at $A$: $y - \frac{b}{\sqrt{2}} = \frac{a}{b} \left( x - \frac{a}{\sqrt{2}} ight) \dots (iii)$.To find the triangle vertices on the $X$-axis,set $y=0$ in $(ii)$ and $(iii)$:For tangent,$0 - \frac{b}{\sqrt{2}} = -\frac{b}{a} (x - \frac{a}{\sqrt{2}}) \implies \frac{a}{\sqrt{2}} = x - \frac{a}{\sqrt{2}} \implies x = \sqrt{2}a$. Point $B = (\sqrt{2}a, 0)$.For normal,$0 - \frac{b}{\sqrt{2}} = \frac{a}{b} (x - \frac{a}{\sqrt{2}}) \implies -\frac{b^2}{\sqrt{2}a} = x - \frac{a}{\sqrt{2}} \implies x = \frac{a}{\sqrt{2}} - \frac{b^2}{\sqrt{2}a} = \frac{a^2-b^2}{\sqrt{2}a}$. Point $C = (\frac{a^2-b^2}{\sqrt{2}a}, 0)$.The height of the triangle is the $y$-coordinate of point $A$,which is $h = \frac{b}{\sqrt{2}}$.The base $BC = \sqrt{2}a - \frac{a^2-b^2}{\sqrt{2}a} = \frac{2a^2 - a^2 + b^2}{\sqrt{2}a} = \frac{a^2+b^2}{\sqrt{2}a}$.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \left( \frac{a^2+b^2}{\sqrt{2}a} ight) 	imes \left( \frac{b}{\sqrt{2}} ight) = \frac{b(a^2+b^2)}{4a}$.

Find the area of the triangle formed by the $X$-axis and the tangent and the normal to the curve $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at the point $\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)$.

Similar Questions

What is the locus of the point of intersection of perpendicular tangents to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$?

The angle between the lines joining the foci of an ellipse to one particular extremity of the minor axis is $90^{\circ}$. The eccentricity of the ellipse is

The sides of the rectangle of greatest area that can be inscribed in the ellipse $x^2+4y^2=64$ are:

For the ellipse $3x^2 + 4y^2 = 12$,the length of the latus rectum is:

Consider the parabola $P : y^2 = 4x$ and the ellipse $E : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Let the line segment joining the points of intersection of $P$ and $E$ be their common latus rectum. If the eccentricity of $E$ is $e$, then $e^2 + 2\sqrt{2}$ is equal to . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module