Consider the parabola y² = 4x. Let S be the focus of the parabola. A pair of tangents drawn to the parabola from the point P = (−2, 1) meet the parabola at P₁ and P₂. Let Q₁ and Q₂ be points on the lines SP₁ and SP₂ respectively such that PQ₁ is perpendicular to SP₁ and PQ₂ is perpendicular to SP₂. Then, which of the following is/are TRUE?

Correct Options:
The correct statements are:
• Q₁Q₂ = $\frac{3\sqrt{10}}{5}$
• PQ₁ = 3
• SQ₂ = 1

Step 1: Identify the focus and tangents to the parabola
Given the equation of the parabola:

$y^2=4x$

Comparing with $y^2=4ax$, we get $a=1$.
Therefore, the focus of the parabola is S = (1, 0).

The equation of a tangent to the parabola in slope form is:

$y=mx+\frac{1}{m}$

Since the tangents pass through the point P = (-2, 1), substitute x = -2 and y = 1:

$1=-2m+\frac{1}{m}$

$2m^2+m-1=0$

$(2m-1)(m+1)=0$

Thus, the slopes of the two tangents are $m_1=\frac{1}{2}$ and $m_2=-1$.

Step 2: Find the points of contact P₁ and P₂
For a tangent with slope m, the point of contact on $y^2=4ax$ is given by $(\frac{a}{m^2},\frac{2a}{m})$.
• For $m_1=\frac{1}{2}$:

${\text{P}}_1=(\frac{1}{{(1/2)}^2},\frac{2}{1/2})=(4,4)$

• For $m_2=-1$:

${\text{P}}_2=(\frac{1}{{(-1)}^2},\frac{2}{-1})=(1,-2)$

Step 3: Equations of lines SP₁ and SP₂ and lengths PQ₁ and SQ₂
The line passing through S(1, 0) and P₁(4, 4) has equation:

$y-0=\frac{4-0}{4-1}(x-1)\Rightarrow 4x-3y-4=0$

Since PQ₁ is perpendicular to SP₁, the distance PQ₁ is the length of the perpendicular from P(-2, 1) to line SP₁:

${\text{PQ}}_1=\frac{\left|4\right(-2)-3(1)-4|}{\sqrt{4^2+{(-3)}^2}}=\frac{|-15|}{5}=3$

Hence, PQ₁ = 3 is TRUE.

Now calculate the distance between point P(-2, 1) and focus S(1, 0):

${\text{SP}}^2={(1-(-2\left)\right)}^2+{(0-1)}^2=3^2+{(-1)}^2=10$

In right-angled triangle PQ₂S (where PQ₂ ⊥ SP₂):

${{\text{SQ}}_2}^2+{{\text{PQ}}_2}^2={\text{SP}}^2$

The line passing through S(1, 0) and P₂(1, -2) is the vertical line $x=1$.
The perpendicular distance from P(-2, 1) to $x=1$ is:

${\text{PQ}}_2=|1-(-2\left)\right|=3$

Therefore, in right triangle PQ₂S:

${{\text{SQ}}_2}^2+3^2=10\Rightarrow {{\text{SQ}}_2}^2=1\Rightarrow {\text{SQ}}_2=1$

Hence, SQ₂ = 1 is TRUE.

Step 4: Find coordinates of Q₁ and Q₂ and the distance Q₁Q₂
Similarly, in right-angled triangle PQ₁S:

${{\text{SQ}}_1}^2+{{\text{PQ}}_1}^2={\text{SP}}^2\Rightarrow {{\text{SQ}}_1}^2+9=10\Rightarrow {\text{SQ}}_1=1$

The unit vector along ray SP₁ (from S(1, 0) towards P₁(4, 4)) is $(\frac{3}{5},\frac{4}{5})$.
Since SQ₁ = 1, the coordinates of Q₁ are:

${\text{Q}}_1=(1+1·\frac{3}{5},0+1·\frac{4}{5})=(\frac{8}{5},\frac{4}{5})$

The ray SP₂ goes vertically downward from S(1, 0). Since SQ₂ = 1, the coordinates of Q₂ are:

${\text{Q}}_2=(1,0-1)=(1,-1)$

Now, calculate distance Q₁Q₂:

${\text{Q}}_1{\text{Q}}_2=\sqrt{{(\frac{8}{5}-1)}^2+{(\frac{4}{5}-(-1\left)\right)}^2}=\sqrt{{\left(\frac{3}{5}\right)}^2+{\left(\frac{9}{5}\right)}^2}=\sqrt{\frac{9+81}{25}}=\frac{\sqrt{90}}{5}=\frac{3\sqrt{10}}{5}$

Hence, Q₁Q₂ = $\frac{3\sqrt{10}}{5}$ is TRUE.