Q1)

first solve both the equations to get point of intersections which are

(-3c+(40-c²)^1/2 /10 , c+3(40-c²)^1/2/10)

(-3c-(40-c²)^1/2/10 , c-3(40-c²)^1/2/10)

then for slope of line given two points on it will be

y₂-y₁/x₂-x₁

therefore m₁=c+3(40-c²)^1/2/-3c+(40-c²)^1/2

m₂=c-3(40-c²)^1/2/-3c-(40-c²)^1/2

use m₁xm₂=-1

then it gives equqtion in c² which is equal to 20

 

hey i am continuing here plz have a look  even at this

(15/AB)²=(3m+1)²/m²+1

(10/AC)²=(2+m)²/1+m²

(6/AD)²=(1-m)²/1+m²

using these values u'll get the value of m=-2/3

therefore equation of line is y+4=m(x+5)

2x+3y+22=0

 

the equation of line is 2x+3y+22=0

solution:

let the equation of line be y+4=m(x+5)

solving x+3y+2=0 and y+4=m(x+5)

we get B=(10-15m/3m+1,3m-4/3m+1)

solving 2x+y+4=0 and y=4=m(x+5)

we get C=(-5m/2+m,6m-8/2+m)

solving x-y-5=0 and y+4=m(x+5)

we get D=(5m+1/1-m,10m-4/1-m)

given condition is (15/AB)²+(10/AC)²=(6/AD)²

equation of  L1 is x/a+y/b=1

mid pt of intercepts of L1 is(a/2,b/2)

equation of L2 is x/p+y/q=1

substitute (a/2,b/2) in L2 as it passes through it

a/2p+b/2q=1

2aq+2bp=4pq

let the acceleration of block A be a1 and block B a2 and the tension in string attached to A be T.

this implies the tension in string attached block B as 2T.

by using virtual work principal  2Tx2=Tx1

2a2=a1

5g-2T=5a2

T=2a1

5g=2T+5a2

5g=13a2

a2=5g/13

a1=10g/13 

hey u have given 2 equations with 3 variables

so we can use the third euation as the point satisfying the circle euation

equation of tangent to y²=4ax is y=mx+a/m

if y=mx+c is a tangent to x²=4by then c=-bm²

a/m=-bm²

m=(-a/b)^1/3

therefore equation of tangent is b^1/3y+a^1/3x+(ab)^2/3=o

i am sorry i took the question wrongly

is it x(4ab)^2/3=2b(y+(4ab)^1/3)

is it x+y=2a

carbon attached alpha carbon is called beta carbon

CH₃-CH₂-OH

allylic carbon is the carbon attached to CH₂=CH-CH₃

hello friend plz check the question once again becoz given conditions doesnot specify a particular ellipse. we well be getting many ellipse

 

slope of AB is +,-root2.

if a normal at t₁ intersects the parabola again at t₂ then t₂=-t₁-2/t₁

as AB subtends a rt angle at (0,0) 2/t₁x2/t₂=-1

substitute t₂ interms of t₁ then value of t₁²==,-2

slope of  AB =2a(t₂-t₁)/a(t₂²-t₁²)

substitute t₂ interms of  t₁

m=-t₁

take the general eqn of circle with center (a,b) and radius r

(x-a)²+(y-b)²=r²

if lx+my+1=0 touches the circle then distance from center to the line is equal to radius

that is (la+mb+1)²=r²(l²+m²)

compare it with given condition which gives 2a=6,2bm=0,a²-r²=4,b²-r²=-5

which gives center to be (3,0)and radius as 5 units.

first write the eqn of chord with midpoint (x₁,y₁) using T=S₁.

YY₁-2A(X+X₁)=(Y₁)²-4AX₁

know compare with the condition for normal that is for a line lx+my+n=0 to be a normal for y2=4ax is al³+2alm²+m²n=0