Mathematics Test - 8

Given parabola can be written as x²=8(y−1/8)

Comparing with standard parabola X2=4aY

X=x, Y=y−1/8, a=2

Therefore Vertex is (X,Y)=(0,0)

⇒(x,y−1/8)=(0,0)

⇒(x,y)=(0,1/8)

Hence option C is the correct answer.

Given: \(h(x)=f(x)-(f(x))^{2}+(f(x))^{3}\)

On differentiating w.r.t. \(x\), we get

\(h^{\prime}(x)=f(x)-2 f(x) \cdot f^{\prime}(x)+3 f^{2}(x) \cdot f^{\prime}(x)\)

\(=f(x)\left[1-2 f(x)+3 f^{2}(x)\right]\)

\(=3 f^{\prime}(x)\left[(f(x))^{2}-\frac{2}{3} f(x)+\frac{1}{3}\right]\)

\(=3 f^{\prime}(x)\left[\left[f(x)-\frac{1}{3}\right]^{2}+\frac{1}{3}-\frac{1}{9}\right]\)

\(=3 f(x)\left[\left[f(x)-\frac{1}{3}\right]^{2}+\frac{3-1}{9}\right]\)

\(=3 f(x)\left[\left[f(x)-\frac{1}{3}\right]^{2}+\frac{2}{9}\right]\)

Note that \(h^{\prime}(x)<0\) if \(f^{\prime}(x)<0\) and \(h^{\prime}(x)>0\)

if \(f^{\prime}(x)>0\)

Therefore, \(h(x)\) is an increasing function if \(f(x)\) is an increasing function, and \(h(x)\) is a decreasing function if \(f(x)\) is a decreasing function.

Therefore, option (A) is correct answer.

Given parabola y₂=4ax

Since, P, S and Q are collinear.

m_PQ=m_PS

2/(t₁+t₂)=2t₁/(t²₁−1)

⇒t²₁−1=t²₁+t₁t₂

⇒t₁t₂=−1

Hence option C is the correct answer.

Therefore, (A) is the answer.

Hence option C is the correct answer.

Length of the double ordinate is parametric form is 8pt=16p

t=2

So coordinates of extremities will be A(8p,8p) and B(8p,−8p)

The slope of the line joining vertex to point A is 1

The slope of the line joining vertex to point B is −1

Product of slopes =−1

Hence the lines are perpendicular

Hence option B is the correct answer.

Given the circle lies in 4^th quadrant, and touches y=0 and x=0 and has radius 3.

∴ Equation of circle is

(x−3)²+(y+3)²=3²

⇒x²+y²−6x+6y+9=0.

Hence option A is the correct answer.

y²+4y+4x+2 = 0

y²+4y+4 = −4x+2

(y+2)²= −4(x−1/2)

So equation of directrix is (x−1/2) = −a = 1 ⇒ x = 3/2

Hence option D is the correct answer.

The given equation of the parabola can be written in the form

(y−3)² = 2(x+2) ⇒ (y−3)²=4.1/2 (c+2)

So, a=1/2

Coordinates of vertex are x+2=0, y−3=0 ⇒ (−2,3)

Coordinates of focus are x+2=1/2, y−3=0 ⇒ (−3/2,3)

Equation of directrix is x+2=−1/2 ⇒ x=−5/2

Hence option D is the correct answer.

The equation of pair of tangent to the circle \(x^{2}+y^{2}+2 g x+2 f y+c=0\) from point \(\left(x_{1}, y_{1}\right)\) is

\(\left(x^{2}+y^{2}+2 g x+2 f y+c\right)\left(x_{1}^{2}+y_{1}^{2}+2 g x_{1}+2 f y_{1}+c\right)\)

\(=\left[x x_{1}+y y_{1}+f\left(x+x_{1}\right)+f\left(y+y_{1}\right)+c\right]^{2}\)

Thus, the equation of pair of tangent to the circle \(x^{2}+y^{2}-2 x-2 h y+h^{2}=0\)

from origin (0,0) is \(\left(x^{2}+y^{2}-2 r x-2 h y+h^{2}\right)\left(0^{2}+0^{2}-2 r 0-2 h 0+h^{2}\right)=\)

\(\left[x 0+y 0-r(x+0)-h(y+0)+h^{2}\right]^{2}\)

\(\Rightarrow\left(x^{2}+y^{2}-2 r x-2 h y+h^{2}\right)\left(h^{2}\right)=\left[-r x-h y+h^{2}\right]^{2}\)

\(\Rightarrow x^{2} h^{2}+h^{2} y^{2}-2 h^{2} r x-2 h^{3} y+h^{4}=\)

\(r^{2} x^{2}+h^{2} y^{2}+h^{4}+2 r x y-2 h^{3} y-2 r x h^{2}\)

\(\Rightarrow h^{2} x^{2}=r^{2} x^{2}+2 r h x y\)

\(\Rightarrow h^{2} x^{2}-r^{2} x^{2}-2 r h y=0\)

\(\Rightarrow x\left(h^{2} x-r^{2} x-2 r h y\right)=0\)

\(\left.\Rightarrow x\left(h^{2}-r^{2}\right) x-2 r h y\right)=0\)

tangent to the circle \(\left(h^{2}-r^{2}\right) x-2 r h y=0, x=0\)

Hence option A is the correct answer.

since, \(B P: A P=3: 1 .\) Then equation of tangent is \(Y-y=f^{\prime}(x)(x-x)\) The intercept on the coordinate axes are \(A\left(x-\frac{y}{f^{\prime}(x)}, 0\right)\)

and \(B[0, y-x f(x)]\)

since, \(P\) is internally intercepts a line \(A B\)

\(\therefore x=\frac{3\left(x-\frac{y}{f^{\prime}(x)}\right)+1 \times 0}{3+1}\)

\(\Rightarrow \frac{d y}{d x}=\frac{y}{-3 x}\)

\(\Rightarrow \frac{d y}{y}=-\frac{1}{3 x} d x\)

On integrating both sides, we get \(x y^{3}=c\) since, curve passes through \((1,1),\) then \(c=1 x y^{3}=1\) At \(x=\frac{1}{8} \Rightarrow y=2\)

Hence, (A) is correct answer.

Given that point P lies on ellipse

it will be in the form of (6√cos θ, 2√sin θ)

The center of ellipse is (0,0) and its distance from P is 2

by using distance formula

⇒6cos²θ+2sin²θ=4

⇒cos²θ=1/2

⇒cosθ=1/√2

therefore the value of θ=π/4

Hence option C is the correct answer.

x²+2y−3x+5=0

⇒x²−3x=−2y−5

⇒x²−2(3/2)x+9/4 = −2y−5+9/4 = −2y−11/4

⇒(x−3/2)² = −2(y−11/4)

Hence latus rectum = 4×1/2=2

Hence option B is the correct answer.

Line passing through (2,−1) is

y+1=m(x−2)

y=mx+(−1−2m)

Condition of tangency is c²=a²m²+b²

(−1−2m)²=3m²+1

1+4m²+4m=3m²+1

m²+4m=0⇒m=0,−4

∴ equation of tangents are y=−1 and 4x+y=7

Hence option B is the correct answer.