Top 50 Most Repeated CONIC SECTIONS PYQs | JEE MAINS
A curated collection of the most important questions from CONIC SECTIONS, fully solved with step-by-step concepts to prepare for JEE MAINS.
A curated collection of the most important questions from CONIC SECTIONS, fully solved with step-by-step concepts to prepare for JEE MAINS.
Substituting $y=mx+c$ into $y^{2}=4ax$ gives a quadratic in $x$. For tangency the discriminant must be zero, leading to $c^{2}=\frac{4a^{2}}{m^{2}}$, ...
Read Full Step-by-Step Solution →Use $e = \sqrt{1 - \frac{b^2}{a^2}}$. Here $a=4$, $b=2$, so $e = \sqrt{1 - \frac{1}{4}} = \sqrt{3}/2$....
Read Full Step-by-Step Solution →Asymptotes: $y = \pm \frac{b}{a}x = \pm \frac{5}{4}x$; slope magnitude = 1.25 → product of slopes = -1.5625 → |slope| = 1.25 → answer = 5/4 = 1.25 → r...
Read Full Step-by-Step Solution →Chord of contact from $(x_1,y_1)$ to circle $x^2+y^2=r^2$ is $xx_1 + yy_1 = r^2$. So $4x + 3y = 25$....
Read Full Step-by-Step Solution →For $y^2 = 4ax$, focus is at (a, 0); here 4a = 12 → a = 3....
Read Full Step-by-Step Solution →Substituting $y=mx+c$ into $y^{2}=4ax$ gives a quadratic in $x$. For tangency the discriminant must be zero, leading to $c^{2}=\frac{4a^{2}}{m^{2}}$, ...
Read Full Step-by-Step Solution →Use $e = \sqrt{1 - \frac{b^2}{a^2}}$. Here $a=4$, $b=2$, so $e = \sqrt{1 - \frac{1}{4}} = \sqrt{3}/2$....
Read Full Step-by-Step Solution →Asymptotes: $y = \pm \frac{b}{a}x = \pm \frac{5}{4}x$; slope magnitude = 1.25 → product of slopes = -1.5625 → |slope| = 1.25 → answer = 5/4 = 1.25 → r...
Read Full Step-by-Step Solution →Chord of contact from $(x_1,y_1)$ to circle $x^2+y^2=r^2$ is $xx_1 + yy_1 = r^2$. So $4x + 3y = 25$....
Read Full Step-by-Step Solution →For $y^2 = 4ax$, focus is at (a, 0); here 4a = 12 → a = 3....
Read Full Step-by-Step Solution →Substituting $y=mx+c$ into $y^{2}=4ax$ gives a quadratic in $x$. For tangency the discriminant must be zero, leading to $c^{2}=\frac{4a^{2}}{m^{2}}$, ...
Read Full Step-by-Step Solution →Use $e = \sqrt{1 - \frac{b^2}{a^2}}$. Here $a=4$, $b=2$, so $e = \sqrt{1 - \frac{1}{4}} = \sqrt{3}/2$....
Read Full Step-by-Step Solution →Asymptotes: $y = \pm \frac{b}{a}x = \pm \frac{5}{4}x$; slope magnitude = 1.25 → product of slopes = -1.5625 → |slope| = 1.25 → answer = 5/4 = 1.25 → r...
Read Full Step-by-Step Solution →Chord of contact from $(x_1,y_1)$ to circle $x^2+y^2=r^2$ is $xx_1 + yy_1 = r^2$. So $4x + 3y = 25$....
Read Full Step-by-Step Solution →For $y^2 = 4ax$, focus is at (a, 0); here 4a = 12 → a = 3....
Read Full Step-by-Step Solution →Substituting $y=mx+c$ into $y^{2}=4ax$ gives a quadratic in $x$. For tangency the discriminant must be zero, leading to $c^{2}=\frac{4a^{2}}{m^{2}}$, ...
Read Full Step-by-Step Solution →Use $e = \sqrt{1 - \frac{b^2}{a^2}}$. Here $a=4$, $b=2$, so $e = \sqrt{1 - \frac{1}{4}} = \sqrt{3}/2$....
Read Full Step-by-Step Solution →Asymptotes: $y = \pm \frac{b}{a}x = \pm \frac{5}{4}x$; slope magnitude = 1.25 → product of slopes = -1.5625 → |slope| = 1.25 → answer = 5/4 = 1.25 → r...
Read Full Step-by-Step Solution →Chord of contact from $(x_1,y_1)$ to circle $x^2+y^2=r^2$ is $xx_1 + yy_1 = r^2$. So $4x + 3y = 25$....
Read Full Step-by-Step Solution →For $y^2 = 4ax$, focus is at (a, 0); here 4a = 12 → a = 3....
Read Full Step-by-Step Solution →Substituting $y=mx+c$ into $y^{2}=4ax$ gives a quadratic in $x$. For tangency the discriminant must be zero, leading to $c^{2}=\frac{4a^{2}}{m^{2}}$, ...
Read Full Step-by-Step Solution →Use $e = \sqrt{1 - \frac{b^2}{a^2}}$. Here $a=4$, $b=2$, so $e = \sqrt{1 - \frac{1}{4}} = \sqrt{3}/2$....
Read Full Step-by-Step Solution →Asymptotes: $y = \pm \frac{b}{a}x = \pm \frac{5}{4}x$; slope magnitude = 1.25 → product of slopes = -1.5625 → |slope| = 1.25 → answer = 5/4 = 1.25 → r...
Read Full Step-by-Step Solution →Chord of contact from $(x_1,y_1)$ to circle $x^2+y^2=r^2$ is $xx_1 + yy_1 = r^2$. So $4x + 3y = 25$....
Read Full Step-by-Step Solution →For $y^2 = 4ax$, focus is at (a, 0); here 4a = 12 → a = 3....
Read Full Step-by-Step Solution →Substituting $y=mx+c$ into $y^{2}=4ax$ gives a quadratic in $x$. For tangency the discriminant must be zero, leading to $c^{2}=\frac{4a^{2}}{m^{2}}$, ...
Read Full Step-by-Step Solution →Use $e = \sqrt{1 - \frac{b^2}{a^2}}$. Here $a=4$, $b=2$, so $e = \sqrt{1 - \frac{1}{4}} = \sqrt{3}/2$....
Read Full Step-by-Step Solution →Asymptotes: $y = \pm \frac{b}{a}x = \pm \frac{5}{4}x$; slope magnitude = 1.25 → product of slopes = -1.5625 → |slope| = 1.25 → answer = 5/4 = 1.25 → r...
Read Full Step-by-Step Solution →Chord of contact from $(x_1,y_1)$ to circle $x^2+y^2=r^2$ is $xx_1 + yy_1 = r^2$. So $4x + 3y = 25$....
Read Full Step-by-Step Solution →For $y^2 = 4ax$, focus is at (a, 0); here 4a = 12 → a = 3....
Read Full Step-by-Step Solution →Substituting $y=mx+c$ into $y^{2}=4ax$ gives a quadratic in $x$. For tangency the discriminant must be zero, leading to $c^{2}=\frac{4a^{2}}{m^{2}}$, ...
Read Full Step-by-Step Solution →Use $e = \sqrt{1 - \frac{b^2}{a^2}}$. Here $a=4$, $b=2$, so $e = \sqrt{1 - \frac{1}{4}} = \sqrt{3}/2$....
Read Full Step-by-Step Solution →Asymptotes: $y = \pm \frac{b}{a}x = \pm \frac{5}{4}x$; slope magnitude = 1.25 → product of slopes = -1.5625 → |slope| = 1.25 → answer = 5/4 = 1.25 → r...
Read Full Step-by-Step Solution →Chord of contact from $(x_1,y_1)$ to circle $x^2+y^2=r^2$ is $xx_1 + yy_1 = r^2$. So $4x + 3y = 25$....
Read Full Step-by-Step Solution →For $y^2 = 4ax$, focus is at (a, 0); here 4a = 12 → a = 3....
Read Full Step-by-Step Solution →