The equation of the line passing through the | vedclass

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

Question

(B) The given line is $yy' = 2ax + 2ax'$,which can be rewritten as $2ax - yy' + 2ax' = 0$.Comparing this with the standard form $Ax + By + C = 0$,the

Vedclass · Accepted Answer

$xy' + 2ay - 2ay' - x'y' = 0$

Vedclass · Answer

(B) The given line is $yy' = 2ax + 2ax'$,which can be rewritten as $2ax - yy' + 2ax' = 0$.Comparing this with the standard form $Ax + By + C = 0$,the slope $m_1 = \frac{2a}{y'}$.Since the required line is perpendicular to this line,its slope $m_2$ must satisfy $m_1 	imes m_2 = -1$.Thus,$m_2 = -\frac{y'}{2a}$.The equation of the line passing through $(x', y')$ with slope $m_2$ is $y - y' = -\frac{y'}{2a}(x - x')$.Multiplying by $2a$,we get $2ay - 2ay' = -y'(x - x')$.$2ay - 2ay' = -xy' + x'y'$.Rearranging the terms,we get $xy' + 2ay - 2ay' - x'y' = 0$.

The equation of the line passing through the point $(x', y')$ and perpendicular to the line $yy' = 2a(x + x')$ is

Similar Questions

The length of the tangent from the point $(4, 5)$ to the circle $x^2 + y^2 + 2x - 6y - 6 = 0$ is

If $P(\frac{\pi}{4})$ and $Q(\frac{\pi}{3})$ are two points on the circle $x^2+y^2-2x-2y-1=0$,then the slope of the tangent to this circle which is parallel to the chord $PQ$ is

The equation of the normal at $t=\frac{\pi}{2}$ to the curve $x=2 \sin t, y=2 \cos t$ is

Consider the following statements:
Assertion $(A)$: The circle $x^2 + y^2 = 1$ has exactly two tangents parallel to the $x$-axis.
Reason $(R)$: $\frac{dy}{dx} = 0$ on the circle exactly at the points $(0, \pm 1)$.
Of these statements:

Find the equation of the circle having normals $(x-1)(y-2)=0$ and a tangent $3x+4y=6$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module