JEE Advanced Level Test: Tangent & Normal- 1 - JEE MCQ

x² + y² = 4
Eqn of tangent at (1,√3)
xx1 + yy1 = 4
x + √3y = 4
√3y = −x + 4
y=−x / √3 + 4/√3
point A will be (4,0)
Eqn of Normal:slope of normal = √3
(y − √3) = √3 (x − 1)
=y − √3 = √3x − √3
y = √3x
Area of triangle = ∫(0 to 1)√3xdx + ∫(1 to 4)(−x/√3) + (4/√3)dx
=√3 [x² / 2] (0 to 1) + [(−x²) /(2√3) + 4x / √3] (1 to 4)
=√3.[1/2 − 0] + [−16 / 2√3 + 16 / √3 + 1/2 √3-4√3] = 8√3 − 7/2√3 + √3/2
=9 / 2√3 + 3 √2 = 12 / 2√3 
= 2√3

y= −(x)^1/2+2....(i)
And bisector of first quadrant is y=x ..... (ii)
On solving Eqs(i) and (ii) we get
x=1,4
∴ From Eq (i)
Points are (1,1) and (4,0)
But (4,0) not satisfy Eq (ii)
∴ Point (1,1) is only point of intersection of curve (i) and line (ii)
Now, slope of tangent of curve (i) is dy/dx = -1/(2(x)1/2)
∴ Slope of tangent at point (1,1) is −1/|dy/dx|(1,1) = -1/2
−1/|dy/dx|(1,1) = -1/(−1/2)
​= 2
∴ Equation of the normal to the curve at point (1,1) will be
y−1=2(x−1)
⇒2x−y−1=0

Slope of such normal is 1.⇒ dy/dx=1
ay² = x³
 ⇒2ay dy/dx = 3x²
⇒ y=(3x²)/2a
 ⇒ a(3x²/2a)²
 =x³
 ⇒x = 4a/9

External tangents are those which touch both the circles but they will not intersect in between the circles. The tangents touch at outmost points of circles that are ends of diameter if the circles have same diameter.

Length of subtangent =y(dx/dy)
Now given √x+√y=3
√y=3−√x
⇒1/2√ydy/dx = −1/(2√x) (On differentiating)
⇒dx/dy=−√x/√y
⇒ydx/dy=−√x/√y
⇒dx/dy/(4,1) = −√1√4
=−2
But the length can never be negative.
So length = 2.

y²=4a[x+asin(x/a)] ..... (i)
∴2y dy/dx​=4a[1+cos(x/a)] ..... (ii)
If tangent is parallel to x-axis, then dy/dx=0
So, from Eq. (i), we get
cos(x/a)=−1
∴sin(x/a)=0
On putting this value in Eq. (i), we get
y²=4a(x+0)
⇒y²=4ax

2x/a²+2y/b²−dy/dx=0
m1=dy/dx=−b²x/a²y
2x/A²−2y/B²⋅dy/dx=0
m2=dy/dx=B²x/A²y
m¹⋅m²=−1
b2/a2⋅x/y⋅B²/A²⋅x/y=1
x²/y²=a²A²/b²B²
x^2/a²+y²/b²−x²/A²+y²/B²=0
x²(1/a²−1/A²)=−y²(1/b²+1/B²)
x²/y²=−(1/b²+1B²)/(1/a²−1/A²)
a²A²/b²B²=−(1/b²+1/B²)(1/a²−1/A²)
a²A²(1/a²−1/A²)=−b²B²(1/b²+1/B²)
a²A²(A²−a²/(a²B²))=−b²B²(B²+b²/(b²B²)
a²−A²=b²+B²
a²−b²=A²+B²