Question

Let $\sf{P(x_1,\:y_1)\:and\:Q(x_2,\:y_2),\:y_1<0,\:y_2<0,}$ be the end points of the latus rectum of the ellipse $\sf{x^2+4y^2=4}$. The equations of parabolas with latus rectum PQ are
$\sf{(a)\:x^2+2\sqrt{3}y=3+\sqrt{3}}$
$\sf{(b)\:x^2-2\sqrt{3}y=3+\sqrt{3}}$
$\sf{(c)\:x^2+2\sqrt{3}y=3-\sqrt{3}}$
$\sf{(d)\:x^2-2\sqrt{3}y=3-\sqrt{3}}$​

Answer 1

First of all let's find out the equation of ellipse .

☆Equation is given as x² + 4y² = 4 .

☆ For converting it into standard form . Divide by 4 both sides .

$\implies \: \frac{ {x}^{2} }{4} \: + \: \frac{ {y}^{2} }{1} \: = \: 1$

Here we got a² = 4 → a = 2

and b² = 1 → b = 1

Now eccentricity will be

$\implies \: e \: = \: \sqrt{1 - { (\frac{b}{a}) }^{2} }$

$e \: = \: \frac{ \sqrt{3} }{2}$

Co ordinate of both P and Q are negative . And end points of latus rectum .So, it's coordinates

P = (- ae , -b/2 )

=( -√3 , -1/2)

Q = ae , -b/2

=( √3, -1/2)

Length of PQ = Latus rectum of parabola .

Length of PQ = 2√3 .

Latus rectum of Parabola = 4a . So ,

☆ 4a = 2√3

☆ a = √3/2

Coordinates of A .

On joining PQ the point where it cuts y axis is focus ( S )

OS = 1/2

S = ( 0 , -1/2 )

AS = a = √3/2

A = ( -1/2 , √3/2)

A' = (-1/2, -√3/2)

A = OS - OA ( Open down wards )

A' = OS + AS ( Open upwards)

Equation → ( x - 0 )² = - 4ay

${x}^{2} \: = \: 4( \frac{ \sqrt{3} }{2} )(y + \frac{1 - \sqrt{3} }{2} )$

x² = - 2√3 y - √3 ( 1- √3)

x² = - 2√3y - √3 + 3 equation 1st

x² +2√3y = 3 - √3

→ ( x - a)² = 4(√3/2)( y - ( -1-√3/2)

→ x² = 2√3y + √3 + 3

→ x² - 2√3y = 3 + √3 equation 2nd

Answer 2

$\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{Question-}}}}}}$

Let $\sf{P(x_1,\:y_1)\:and\:Q(x_2,\:y_2),\:y_1<0,\:y_2<0,}$ be the end points of the latus rectum of the ellipse $\sf{x^2+4y^2=4}$. The equations of parabolas with latus rectum PQ are

$\sf{(a)\:x^2+2\sqrt{3}y=3+\sqrt{3}}$

$\sf{(b)\:x^2-2\sqrt{3}y=3+\sqrt{3}}$

$\sf{(c)\:x^2+2\sqrt{3}y=3-\sqrt{3}}$

$\sf{(d)\:x^2-2\sqrt{3}y=3-\sqrt{3}}$

$\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{Answer:-}}}}}}$

Equation is x² + 4y² = 4 ,For converting it into standard form Divide by 4 both sides

$\begin{lgathered}\implies \: \frac{ {x}^{2} }{4} \: + \: \frac{ {y}^{2} }{1} \: = \: 1 \\\end{lgathered}$

we get a² = 4 → a = 2

and b² = 1 → b = 1

eccentricity -

$\begin{lgathered}\implies \: e \: = \: \sqrt{1 - { (\frac{b}{a}) }^{2} } \\\end{lgathered}$

$\begin{lgathered}e \: = \: \frac{ \sqrt{3} }{2} \\\end{lgathered}$

Co ordinate of P and Q are -ve. And end points of latus rectum,cordinates-

P = (- ae , -b/2 )

=( -√3 , -1/2)

Q = ae , -b/2

=( √3, -1/2)

Length of PQ = Latus rectum of parabola

Length of PQ = 2√3 .

Latus rectum of Parabola = 4a . So ,

4a = 2√3

a = √3/2

Coordinates of A .

On joining PQ the point where it cuts y axis is focus ( S )

OS = 1/2

S = ( 0 , -1/2 )

AS = a = √3/2

A = ( -1/2 , √3/2)

A' = (-1/2, -√3/2)

A = OS - OA ( Open down wards )

A' = OS + AS ( Open upwards)

Equation ⇝ ( x - 0 )² = - 4ay

$\begin{lgathered}{x}^{2} \: = \: 4( \frac{ \sqrt{3} }{2} )(y + \frac{1 - \sqrt{3} }{2} ) \\\end{lgathered}$

x² = - 2√3 y - √3 ( 1- √3)

⇝x² = - 2√3y - √3 + 3 (equation 1st)

x² +2√3y = 3 - √3

⇝ ( x - a)² = 4(√3/2)( y - ( -1-√3/2)

⇝ x² = 2√3y + √3 + 3

⇝ x² - 2√3y = 3 + √3 ( equation 2nd)