Question

The length of the major axis of the ellipse is $(5x-10)²+(5+15)²=$$\dfrac{(3x-4y+7)²}{4}$​

Answer 1

Answer:

(5x−10) 2

+(5y+15) 2

=4

(3x−4y+7) 2

(x−2)

2+(y+3) 2

=( 2 153x−4y−7 ) 2

(x−2)

2 +(y+3) 2

= 21

5 ∣3x−4y−7∣

is an ellipse , whose focus is (2,-3), directrix 3x-4y+7 =0 and eccentricity is 21

Length of perpendicular from focus to directrix i 5

∣3×2−4(−3)+7∣

=5ea−ae=5

2a− 2a=5

a= 310

So length of major axis is

320

hope it help.

mark brainliest

Answer 2

Appropriate Question is

The length of major axis of the ellipse is

$\rm :\:{(5x - 10)}^{2}+ {(5y + 15)}^{2}=\dfrac{{(3x - 4y + 7)}^{2} }{4}$

Basic Concept Used :-

Definition of ellipse :-

The set of locus of point P which moves in such a way that it distance from the fixed point (S) and perpendicular distance on the fixed line always bear a constant ratio which is always less than 1.

• The fixed point is called focus.

• The fixed line is called directrix.

• The constant ratio is called eccentricity represents by e.

$\large\underline{\bf{Solution-}}$

Given

$\rm :\longmapsto\:{(5x - 10)}^{2}+ {(5y + 15)}^{2}=\dfrac{{(3x - 4y + 7)}^{2} }{4}$

$\rm :\longmapsto\: 25{(x - 2)}^{2} + 25{(y + 3)}^{2} = \dfrac{ {(3x - 4y + 7)}^{2} }{4}$

$\rm :\longmapsto\:{(x - 2)}^{2}+ {(y + 3)}^{2}=\dfrac{{(3x - 4y + 7)}^{2} }{4 \times 25}$

$\rm :\longmapsto\:{(x - 2)}^{2}+ {(y + 3)}^{2}=\dfrac{{(3x - 4y + 7)}^{2} }{ {2}^{2} \times {5}^{2} }$

$\rm \therefore \: \sqrt{ {(x - 2)}^{2} + {(y + 3)}^{2}} = \dfrac{1}{2} \times \dfrac{ |3x - 4y + 7| }{5}$

So, by definition of ellipse,

its provides

• focus is (2, - 3)

• Equation of directrix is 3x - 4y + 7 = 0

• Eccentricity (e) = 1/2.

We know that,

The perpendicular distance (d) drawn from a point (p, q) on the line ax + by + c = 0 is

$\bf \:Distance,d = \dfrac{ |ap + bq + c| }{ \sqrt{ {a}^{2} + {b}^{2} } }$

Now,

Distance (d) of the directrix 3x - 4y + 7 = 0 from point (2, - 3) is given by

$\rm :\longmapsto\:\bf \:Distance,d = \dfrac{ |3 \times 2 + 4 \times 3 + 7| }{ \sqrt{ {3}^{2} + {4}^{2} } }$

$\rm :\longmapsto\:\bf \:Distance,d = \dfrac{ |6 + 12 + 7| }{ \sqrt{9 + 16 } }$

$\rm :\longmapsto\:\bf \:Distance,d = \dfrac{ |25| }{ \sqrt{25} }$

$\rm :\longmapsto\:\bf \:Distance,d = 5 \: units$

Now, we know that,

$\sf \: Distance \: of \: focus \: from \: centre \: = ae$

and

$\sf \: Distance \: of \: directrix \: from \: centre\: = \dfrac{a}{e}$

Hence,

$\rm :\longmapsto\:\dfrac{a}{e} - ae = 5$

$\rm :\longmapsto\:a \bigg(\dfrac{1}{e} - e \bigg) = 5$

$\rm :\longmapsto\:a \bigg(2 - \dfrac{1}{2} \bigg) = 5 \: \: \: \: \: \: \: \{ \: \because \: e = \dfrac{1}{2} \}$

$\rm :\longmapsto\:a \bigg(\dfrac{4 - 1}{2} \bigg) = 5$

$\rm :\longmapsto\:a \bigg(\dfrac{3}{2} \bigg) = 5$

$\bf\implies \:a \: = \: \dfrac{10}{3}$

Now,

We know

$\rm :\longmapsto\:Length \: of \: major \: axis \: = 2a$

$\rm :\longmapsto\:Length \: of \: major \: axis \: = 2 \times \dfrac{10}{3}$

$\rm :\longmapsto\:Length \: of \: major \: axis \: = \dfrac{20}{3} \: units$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \underbrace{ \purple{\boxed{\bf \: Length \: of \: major \: axis \: = \dfrac{20}{3} \: units}}}$

Additional Information :-

The general equation of ellipse having centre at (0, 0) is

$\rm :\longmapsto\:\dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } = 1 \: where \: a \: \ne \: \: b$

$\boxed{\bf \: Vertex = ( \pm \: a,0)}$

$\boxed{\bf \: eccentricity, \: e = \sqrt{1 - \dfrac{ {b}^{2} }{ {a}^{2}}}}$

$\boxed{\bf \: {b}^{2} = {a}^{2}(1 - {e}^{2}}$

$\boxed{\bf \:Focus = ( \pm \: ae, \: 0)}$

$\boxed{\bf \: Length \: of \: minor \: axis = 2b}$

$\boxed{\bf \: Length \: of \: major \: axis = 2a}$

$\boxed{\bf \: Length \: of \: latus\: rectum = \dfrac{ {2b}^{2} }{a} }$