The pole of the straight line x + 4y = 4 with | vedclass

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

Question

(B) The equation of the ellipse is $x^2 + 4y^2 = 4$,which can be written as $\frac{x^2}{4} + \frac{y^2}{1} = 1$.Here,$a^2 = 4$ and $b^2 = 1$.Let the p

Vedclass · Accepted Answer

$(1, 1)$

Vedclass · Answer

(B) The equation of the ellipse is $x^2 + 4y^2 = 4$,which can be written as $\frac{x^2}{4} + \frac{y^2}{1} = 1$.Here,$a^2 = 4$ and $b^2 = 1$.Let the pole be $(h, k)$. The equation of the polar of the point $(h, k)$ with respect to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $\frac{hx}{a^2} + \frac{ky}{b^2} = 1$.Substituting the values,we get $\frac{hx}{4} + \frac{ky}{1} = 1$,which simplifies to $hx + 4ky = 4$.Comparing this with the given line $x + 4y = 4$,we get $h = 1$ and $k = 1$.Therefore,the pole is $(1, 1)$.

The pole of the straight line $x + 4y = 4$ with respect to the ellipse $x^2 + 4y^2 = 4$ is

Similar Questions

If the slope of one line of the pair of lines $2x^2 + hxy + 6y^2 = 0$ is thrice the slope of the other line,then $h =$

An ovule which becomes curved so that the nucellus and embryo sac lie at right angles to the funicle is:

$A$ conducting circular loop is placed in a uniform magnetic field,$B = 0.025 \, T$,with its plane perpendicular to the magnetic field. The radius of the loop is made to shrink at a constant rate of $1 \, mm/s$. The induced $e.m.f.$ when the radius is $2 \, cm$ is

The catalyst and promoter respectively used in the Haber's process of industrial synthesis of ammonia are

$x = \log \left( \frac{1}{y} + \sqrt{1 + \frac{1}{y^2}} \right) \Rightarrow y$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module