Question

Find the possible values of k for which the line y = 1/2x + k is a tangent to the curve x(square) +y(square)=8k.​

Answer 1

Ways to Solve

Basis- Triangle(Sol. 1)

• Triangles

→ The area of a triangle is used.

• Linear Equation

→ Intercept form $\dfrac{x}{a} + \dfrac{y}{b} = 1$

Basis- Distance(Sol. 2)

• Property of Tangent

→ The distance between the circumcentre and the tangent is the radius.

• Linear Equation

→ General form $ax+by+c=0$

Solution

Sol. 1

(Attachment is included!!!)

Let us rewrite the linear equation.

$y = \dfrac{1}{2} x + k$

$\Leftrightarrow -\dfrac{1}{2} x + y = k$

$\Leftrightarrow \dfrac{x}{-2k} + \dfrac{y}{k} = 1$

Thus, it is a line that x and y-intercepts are $-2k$ and $k$ respectively. Now, the bounded area between the line and the axes is a right triangle.

• Length of the hypotenuse $\sqrt{5} k$
• Two sides $|-2k|$ and $|k|$
• Perpendicular down the hypotenuse $\sqrt{8k}$

The area of the triangle is half the base×height. Multiply to 2 and we remove the fraction.

$|-2k|\times |k|=\sqrt{5} k\times \sqrt{8k}$

$2k^2=\sqrt{40k^3}$

$4k^4=40k^3$

$4k^3(k-10)=0$

Thus, the required solution is $k=10$.

Sol. 2

$d = \dfrac{|2k|}{\sqrt{1+4} }$ is the distance $d$ between $x-2y+2k=0$, and $(0,0)$.

Based on the property of tangent, the distance is equal to the radius.

$\sqrt{8k}=\dfrac{|2k|}{\sqrt{5} }$

$8k=\dfrac{4k^2}{5}$

$4k^2-40k=0$

$4k(k-10)=0$

Thus, the required solution is $k=10$.

Answer 2

Answer:

Ways to Solve

Basis- Triangle(Sol. 1)

Triangles

→ The area of a triangle is used.

Linear Equation

→ Intercept form \dfrac{x}{a} + \dfrac{y}{b} = 1

a

x

+

b

y

=1

Basis- Distance(Sol. 2)

Property of Tangent

→ The distance between the circumcentre and the tangent is the radius.

Linear Equation

→ General form ax+by+c=0ax+by+c=0

Solution

Sol. 1

(Attachment is included!!!)

Let us rewrite the linear equation.

y = \dfrac{1}{2} x + ky=

2

1

x+k

\Leftrightarrow -\dfrac{1}{2} x + y = k⇔−

2

1

x+y=k

\Leftrightarrow \dfrac{x}{-2k} + \dfrac{y}{k} = 1⇔

−2k

x

+

k

y

=1

Thus, it is a line that x and y-intercepts are -2k−2k and kk respectively. Now, the bounded area between the line and the axes is a right triangle.

Length of the hypotenuse \sqrt{5} k

5

k

Two sides |-2k|∣−2k∣ and |k|∣k∣

Perpendicular down the hypotenuse \sqrt{8k}

8k

The area of the triangle is half the base×height. Multiply to 2 and we remove the fraction.

|-2k|\times |k|=\sqrt{5} k\times \sqrt{8k}∣−2k∣×∣k∣=

5

k×

8k

2k^2=\sqrt{40k^3}2k

2

=

40k

3

4k^4=40k^34k

4

=40k

3

4k^3(k-10)=04k

3

(k−10)=0

Thus, the required solution is k=10k=10 .

Sol. 2

d = \dfrac{|2k|}{\sqrt{1+4} }d=

1+4

∣2k∣

is the distance dd between x-2y+2k=0x−2y+2k=0 , and (0,0)(0,0) .

Based on the property of tangent, the distance is equal to the radius.

\sqrt{8k}=\dfrac{|2k|}{\sqrt{5} }

8k

=

5

∣2k∣

8k=\dfrac{4k^2}{5}8k=

5

4k

2

4k^2-40k=04k

2

−40k=0

4k(k-10)=04k(k−10)=0

Thus, the required solution is k=10k=10